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CHEMISTRY, M3 EQ-Bank 20

During a laboratory investigation, a student mixed two solutions and observed a sudden colour change, an increase in temperature, and the formation of bubbles.

  1. Explain why these observations indicate a chemical change.   (3 marks)

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  1. Describe two types of chemical reactions that could cause at least two of these observations each.   (2 marks)

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Show Answers Only

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

 

Show Worked Solution

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

Filed Under: Chemical Reactions Tagged With: Band 5, Band 6, smc-4263-15-Predicting products, smc-4263-8-Indicators of chemical change

CHEMISTRY, M3 EQ-Bank 6 MC

Which of the following correctly identifies the gas or gases released when hydrochloric acid reacts with magnesium, potassium hydroxide, calcium carbonate, and ammonium carbonate, respectively?

\begin{align*}
\begin{array}{l}
\ & \\
\ & \\
\textbf{A.}\\
\textbf{}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|l|l|l|l|}
\hline
\textit{Magnesium} & \textit{Potassium} & \textit{Calcium} & \textit{Ammonium} \\
& \textit{hydroxide} & \textit{carbonate} & \textit{carbonate} \\
\hline
\text{No gas} & \text{Hydrogen} & \text{Carbon dioxide} & \text{Carbon dioxide} \\
& & & \text{and ammonia} \\
\hline
\text{Hydrogen} & \text{No gas} & \text{Carbon dioxide} & \text{Ammonia} \\
\hline
\text{Carbon dioxide} & \text{Hydrogen} & \text{No Gas} & \text{Carbon dioxide} \\
\hline
\text{Hydrogen} & \text{No gas} & \text{Carbon Dioxide} & \text{Carbon dioxide} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution
  • \(\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}\)  (Hydrogen gas produced)
  • \(\ce{KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)}\)  (No gas produced)
  • \(\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}\)  (Carbon dioxide produced)
  • \(\ce{(NH4)2CO3(s) + 2HCl(aq) -> 2NH4Cl(aq) + CO2(g) + H2O(l)}\)  (Carbon dioxide produced)

\(\Rightarrow D\)

Filed Under: Chemical Reactions Tagged With: Band 5, smc-4263-15-Predicting products

CHEMISTRY, M3 EQ-Bank 18

A student heats sodium metal, copper carbonate and propane gas \(\ce{(C3H8)}\) individually with a Bunsen burner. All of the substances react but only two of the substances react with the oxygen in the air.

Write a balanced chemical equation for each of the reactions that occurred.   (3 marks)

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Combustion of sodium metal:  \(\ce{2Na(s) + O2(g) -> 2NaO(s)}\)

Combustion of propane gas:  \(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}\)

Decomposition of copper carbonate:  \(\ce{CuCO3(s) -> CuO(s) + CO2(g)}\)

Show Worked Solution

Combustion of sodium metal:  \(\ce{2Na(s) + O2(g) -> 2NaO(s)}\)

Combustion of propane gas:  \(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}\)

Decomposition of copper carbonate:  \(\ce{CuCO3(s) -> CuO(s) + CO2(g)}\)

  • The last reaction is endothermic and so requires the heat of the Bunsen burner to proceed where as the first two require the heat of the Bunsen burner to overcome their activation energies.

Filed Under: Chemical Reactions Tagged With: Band 5, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-30-Combustion

CHEMISTRY, M3 EQ-Bank 17

A student tested how soluble silver salts are by reacting a 0.1 mol L\(^{-1}\) silver nitrate solution with 0.1 mol L\(^{-1}\) solutions of calcium hydroxide, calcium chloride, and calcium sulfate. The results are shown below:

\begin{array} {|l|l|}
\hline \ \ \ \ \ \text{Compound} & \ \ \ \ \ \text{Observation} \\
\hline \text{calcium hydroxide} & \text{No reaction} \\
\hline \text{calcium chloride} & \text{White precipitate} \\
\hline \text{calcium sulfate} & \text{No reaction} \\
\hline \end{array}

  1. Write a balanced chemical equation for the reaction with calcium chloride.   (2 marks)

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  1. Name the white precipitate.   (1 mark)

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a.    \(\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}\)

b.    The white precipitate is \(\ce{AgCl}\) → silver chloride.

Show Worked Solution

a.    \(\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}\)
 

b.    The white precipitate is \(\ce{AgCl}\) → silver chloride.

Filed Under: Chemical Reactions Tagged With: Band 3, Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-25-Precipitation

CHEMISTRY, M3 EQ-Bank 28v4

A student stirs 2.50 g of silver (I) nitrate powder into 100.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

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    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

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a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(2.11 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(\ce{n(AgNO3) = \frac{m}{M} = \frac{2.50}{107.9 + 14.01 + 16.00 \times 3} = \frac{2.50}{169.91} = 0.01471 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.100 = 0.150 \text{ mol}}\)

\(\ce{AgNO3} \text{ is the limiting reagent}\)

\(\ce{n(AgCl) = n(AgNO3) = 0.01471 \text{ mol}}\)

\(\ce{m(AgCl) = n \times M = 0.01471 \times (107.9 + 35.45) = 0.01471 \times 143.35 = 2.11 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Filed Under: Chemical Reactions - draft Tagged With: Band 4, Band 5, Band 6, smc-4263-15-Predicting products, smc-4263-25-Precipitation

CHEMISTRY, M3 EQ-Bank 29v4

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between aluminium and nitric acid.   (1 mark)

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  2. The decomposition of mercury (II) oxide.    (1 mark)

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  3. The complete combustion of pentane (\(\ce{C5H12}\)) with excess oxygen.   (1 mark)

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  4. The reaction between magnesium hydroxide and acetic acid (\(\ce{CH3COOH}\)).   (1 mark)

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  5. The reaction between calcium carbonate and hydrochloric acid.   (1 mark)

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a.    \(\ce{2Al + 6HNO3 -> 2Al(NO3)3 + 3H2}\)

b.    \(\ce{2HgO -> 2Hg + O2}\)

c.    \(\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}\)

d.    \(\ce{Mg(OH)2 + 2CH3COOH -> Mg(CH3COO)2 + 2H2O}\)

e.    \(\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}\)

Show Worked Solution

a.    Active metal and acid

\(\ce{2Al + 6HNO3 -> 2Al(NO3)3 + 3H2}\)

b.    Decomposition 

\(\ce{2HgO -> 2Hg + O2}\)

c.    Combustion

\(\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}\)

d.    Acid-base

\(\ce{Mg(OH)2 + 2CH3COOH -> Mg(CH3COO)2 + 2H2O}\)

e.    Acid-carbonate

\(\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}\)

Filed Under: Chemical Reactions - draft Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products

CHEMISTRY, M3 EQ-Bank 29v3

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between iron and sulfuric acid.   (1 mark)

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  2. The decomposition of sodium carbonate.   (1 mark)

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  3. The incomplete combustion of propene (\(\ce{C3H6}\)) with a limited amount of oxygen.   (1 mark)

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  4. The reaction between calcium hydroxide and phosphoric acid.   (1 mark)

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  5. The reaction between sodium carbonate and nitric acid.   (1 mark) 

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a.    \(\ce{Fe + H2SO4 -> FeSO4 + H2}\)

b.    \(\ce{Na2CO3 -> Na2O + CO2}\)

c.    \(\ce{2C3H6 + 9O2 -> 6CO + 6H2O}\)

d.    \(\ce{3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O}\)

e.    \(\ce{Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O}\)

Show Worked Solution

a.    Active metal and acid

\[\ce{Fe + H2SO4 -> FeSO4 + H2}\]

b.    Decomposition

\[\ce{Na2CO3 -> Na2O + CO2}\]

c.   Combustion

\[\ce{2C3H6 + 9O2 -> 6CO + 6H2O}\]

d.    Acid-Base

\[\ce{3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O}\]

e.    Acid-Carbonate

\[\ce{Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O}\]

Filed Under: Chemical Reactions - draft Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products

CHEMISTRY, M3 EQ-Bank 28v2

A student stirs 3.50 g of copper (II) nitrate powder into 150.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

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  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

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    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

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a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(2.51 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(\ce{n(Cu(NO3)2) = \frac{m}{M} = \frac{3.50}{63.55 + 2 \times (14.01 + 16.00 \times 3)} = \frac{3.50}{187.57} = 0.01866 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.150 = 0.225 \text{ mol}}\)

\(\ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCl2) = n(Cu(NO3)2) = 0.01866 \text{ mol}}\)

\(\ce{m(CuCl2) = n \times M = 0.01866 \times (63.55 + 2 \times 35.45) = 0.01866 \times 134.45 = 2.51 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Filed Under: Chemical Reactions - draft Tagged With: Band 4, Band 5, Band 6, smc-4263-15-Predicting products, smc-4263-25-Precipitation

CHEMISTRY, M3 EQ-Bank 16

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between lithium hydroxide and hydrochloric acid.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. The reaction between potassium carbonate and hydrochloric acid.   (1 mark)

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a.    \(\ce{LiOH + HCl -> LiCl + H2O}\)

b.    \(\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\)

Show Worked Solution

a.    Acid-Base

\[\ce{LiOH + HCl -> LiCl + H2O}\]

b.    Acid-carbonate

\[\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\]

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-40-Acid/base reactions

CHEMISTRY, M3 EQ-Bank 15

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between zinc and sulfuric acid.   (1 mark)

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  2. The decomposition of calcium carbonate.   (1 mark)

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  3. The incomplete combustion of ethane \(\ce{(C2H6)}\) with a limited amount of oxygen.   (1 mark)

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a.    \(\ce{Zn + H2SO4 -> ZnSO4 + H2}\)

b.    \(\ce{CaCO3 -> CaO + CO2}\)

c.    \(\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\)

Show Worked Solution

a.    Active metal and acid

\[\ce{Zn + H2SO4 -> ZnSO4 + H2}\]

b.    Decomposition

\[\ce{CaCO3 -> CaO + CO2}\]

c.    Combustion

\[\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\]

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-30-Combustion

CHEMISTRY, M3 EQ-Bank 14

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between potassium hydroxide and sulfuric acid.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. The reaction between sodium bicarbonate \(\ce{(NaHCO3)}\) and acetic acid \(\ce{(CH3COOH)}\).   (1 mark)

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a.    \(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.    \(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

Show Worked Solution

a.   Acid-base 

\(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.   Acid-carbonate 

\(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-40-Acid/base reactions

CHEMISTRY, M3 EQ-Bank 13

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between magnesium and hydrochloric acid.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. The decomposition of ammonia.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  3. The incomplete combustion of butane (\(\ce{C4H10}\)) with a limited amount of oxygen.   (1 mark)

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a.    \(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.    \(\ce{2NH3 -> N2 + 3H2}\)

c.    \(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

Show Worked Solution

a.   Reaction between active metal and acid

\(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.   Decomposition

\(\ce{2NH3 -> N2 + 3H2}\)

c.   Combustion

\(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-30-Combustion

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

    --- 8 WORK AREA LINES (style=lined) ---

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)

b.    \(2.06 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
 

b.   \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)

\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)

\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)

\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)

\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

Filed Under: Chemical Reactions Tagged With: Band 4, Band 5, Band 6, smc-4263-15-Predicting products, smc-4263-25-Precipitation

CHEMISTRY, M3 EQ-Bank 11

A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

    --- 6 WORK AREA LINES (style=lined) ---

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

    --- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)

b.    \(1.61 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
 

b.   \(\ce{MM(Cu(NO3)2) = 63.55 + 2 \times (14.01 + 16.00 \times 3) = 187.57}\)

\(\ce{n(Cu(NO3)2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.45}{187.57} = 0.01306 \text{ mol}}\)

\(\ce{n(Na2CO3) = c \times V = 1.25 \times 0.200 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCO3) = n(Cu(NO3)2) = 0.01306 \text{ mol}}\)

\(\ce{m(CuCO3) = n \times MM = 0.01306 \times (63.55 + 12.01 + 16.00 \times 3) = 1.61 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

Filed Under: Chemical Reactions Tagged With: Band 4, Band 5, Band 6, smc-4263-15-Predicting products, smc-4263-25-Precipitation

CHEMISTRY, M3 SM-Bank 12 MC

Electrolysis and photolysis are examples of which type of reaction? 

  1. Combustion
  2. Acid/Base 
  3. Synthesis
  4. Decomposition 
Show Answers Only

\(D\)

Show Worked Solution
  • Both electrolysis and photolysis involve the breakdown of compounds into simpler substances under decomposition reactions.

\(\Rightarrow D\)

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-15-Predicting products

CHEMISTRY, M3 EQ-Bank 9

Two moles of butane \(\ce{C3H8(g)}\) were reacted with 224 grams of oxygen \(\ce{O2(g)}\).

  1. Write the balanced equation for this reaction.   (3 marks)

    --- 6 WORK AREA LINES (style=lined) ---

  2. Determine the mass, in grams, of \(\ce{CO2(g)}\) produced by this reaction.   (1 mark)

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a.    \(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)

b.    \(\ce{m(CO2) = 88.02\ \text{g}}\)

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a.    Complete combustion equation:

   \(\ce{C3H8(g) + 5O2(g)\ \rightarrow 3CO2(g) + 4H2O(l)} \)

  • Two moles of butane require 10 moles of oxygen to fully combust. 
  •    \(\ce{n(O2) = \dfrac {m}{MM}= \dfrac {224}{32} = 7}\)
  • Oxygen is limiting reagent and butane will undergo incomplete combustion according to the following balanced equation:
  •    \(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)

b.    Using the equation in part (i), 2 moles of \(\ce{CO2}\) will be produced

   \(\ce{m(CO2) = n \times MM = 2 \times 44.01 = 88.02\ \text{g}}\)

Filed Under: Chemical Reactions Tagged With: Band 4, Band 5, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-30-Combustion

CHEMISTRY, M3 EQ-Bank 7

Sodium and zinc are each reacted with water and steam. Predict the observations that would be made about these reactions.   (4 marks)

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Sodium \((\ce{Na}) \):

  • Sodium is an alkali metal that will react vigorously with both water and steam to produce a hydroxide and hydrogen gas according to the chemical equation
  •    \(\ce{2Na(s) + 2H2O(l) \rightarrow 2NaOH(aq) + H2(g)} \)
  • The reaction will show no colour change in the aqueous solution as \(\ce{Na(aq)}\) is colourless like water.

Zinc \((\ce{Zn})\):

  • Zinc is less reactive and will not react with water but will form an oxide with steam (due to the higher energy present) according to the equation
  •    \(\ce{Zn(s) + H2O(steam) \rightarrow ZnO(aq) + H2(g)} \)
  • While zinc metal is a silver colour, zinc oxide is white.
Show Worked Solution

Sodium \((\ce{Na}) \):

  • Sodium is an alkali metal that will react vigorously with both water and steam to produce a hydroxide and hydrogen gas according to the chemical equation
  •    \(\ce{2Na(s) + 2H2O(l) \rightarrow 2NaOH(aq) + H2(g)} \)
  • The reaction will show no colour change in the aqueous solution as \(\ce{Na(aq)}\) is colourless like water.

Zinc \((\ce{Zn})\):

  • Zinc is less reactive and will not react with water but will form an oxide with steam (due to the higher energy present) according to the equation
  •    \(\ce{Zn(s) + H2O(steam) \rightarrow ZnO(aq) + H2(g)} \)
  • While zinc metal is a silver colour, zinc oxide is white.

Filed Under: Chemical Reactions Tagged With: Band 4, smc-4263-15-Predicting products, smc-4263-8-Indicators of chemical change

CHEMISTRY, M3 EQ-Bank 6

Three unknown metals are reacted with dilute \(\ce{HCl(aq)}\) and the following observations are made:  

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \textit{Metal} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \textit{Observations} \\
\hline
\rule{0pt}{2.5ex} \text{A} \rule[-1ex]{0pt}{0pt} & \text{No observable reaction} \\
\hline
\rule{0pt}{2.5ex} \text{B} \rule[-1ex]{0pt}{0pt} & \text{Slow bubbling} \\
\hline
\rule{0pt}{2.5ex} \text{C} \rule[-1ex]{0pt}{0pt} &  \text{Fast, abrupt bubbling} \\
\hline
\end{array}

You are told that the metals in question are Magnesium, Platinum and Zinc.

  1. Explain which of the above metals correspond to \(\text{A}\), \(\text{B}\) and \(\text{C}\), giving reasons for your answer.   (3 marks) 

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  2. Describe the type of reaction between \(\ce{HCl}\) and metal \(\ce{C}\).   (1 mark)

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  3. Provide one balanced chemical equation between \(\ce{HCl}\) and either metal \(\text{B}\) or metal \(\text{C}\).   (1 mark)

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a.    Identifying metals \(\text{A}\), \(\text{B}\) and \(\text{C}\):

  • The metals \(\text{A}\), \(\text{B}\) and \(\text{C}\) differ by their relative reactivity.
  • Magnesium is the most reactive as it is an alkali earth metal (Metal \(\text{C}\)).
  • Zinc and Platinum are both less reactive than magnesium. Zinc however exists higher on the reactivity series of metals than Platinum which is the least reactive of the three metals.
  • Therefore, metal \(\text{B}\) is zinc and metal \(\text{A}\) is platinum. 

b.   A single displacement reaction.
 

c.    Include one of the following:

\(\ce{Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g) }\)

\(\ce{Zn(s) + 2HCl(aq) \rightarrow ZnCl2(aq) + H2(g) }\)

Show Worked Solution

a.    Identifying metals \(\text{A}\), \(\text{B}\) and \(\text{C}\):

  • The metals \(\text{A}\), \(\text{B}\) and \(\text{C}\) differ by their relative reactivity.
  • Magnesium is the most reactive as it is an alkali earth metal (Metal \(\text{C}\)).
  • Zinc and Platinum are both less reactive than magnesium. Zinc however exists higher on the reactivity series of metals than Platinum which is the least reactive of the three metals.
  • Therefore, metal \(\text{B}\) is zinc and metal \(\text{A}\) is platinum. 

b.   A single displacement reaction.
 

c.    Include one of the following:

\(\ce{Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g) }\)

\(\ce{Zn(s) + 2HCl(aq) \rightarrow ZnCl2(aq) + H2(g) }\)

Filed Under: Chemical Reactions Tagged With: Band 4, Band 5, smc-4263-10-Balanced chemical equations, smc-4263-15-Predicting products, smc-4263-8-Indicators of chemical change

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