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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 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)
  2. The decomposition of mercury (II) oxide.    (1 mark)
  3. The complete combustion of pentane (\(\ce{C5H12}\)) with excess oxygen.   (1 mark)
  4. The reaction between magnesium hydroxide and acetic acid (\(\ce{CH3COOH}\)).   (1 mark)
  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

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

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)
  2. The decomposition of sodium carbonate.   (1 mark)
  3. The incomplete combustion of propene (\(\ce{C3H6}\)) with a limited amount of oxygen.   (1 mark)
  4. The reaction between calcium hydroxide and phosphoric acid.   (1 mark)
  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

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

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.