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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)
  1. Name the white precipitate.   (1 mark)
Show Answers Only

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.

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

    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)
Show Answers Only

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.

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

    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)
Show Answers Only

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.

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

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)

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.

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)

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

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)

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.

CHEMISTRY, M3 EQ-Bank 4

When aqueous lead \(\text{(II)}\) nitrate \(\ce{(Pb(NO3)2)}\) is mixed with a potassium iodide solution \((\ce{KI}) \), a precipitation of lead \(\text{(II)}\) iodide \(\ce{(PbI2)}\) results.

Write a balanced chemical equation for this precipitation reaction.  (2 marks)

Show Answers Only

\(\ce{Pb(NO3)2(aq) + 2KI(aq) \rightarrow PbI2(s) + 2KNO3(aq)}\)

Show Worked Solution

→ Reaction products: \(\ce{PbI2,\ KNO3}\)

→ Balanced equation:

\(\ce{Pb(NO3)2(aq) + 2KI(aq) \rightarrow PbI2(s) + 2KNO3(aq)}\)

CHEMISTRY, M3 EQ-Bank 1

Complete the table below by outlining one physical indication of each chemical change described below.   (3 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication}\ \ \ \ \ \ \ \  \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \text{Formation of gas (bubbles)} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \text{Formation of gas (bubbles)} \\
\hline
\end{array}