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CHEMISTRY, M2 EQ-Bank 8

A small community pool generates chlorine gas on site by reacting manganese dioxide \(\ce{(MnO2)}\) with hydrochloric acid. The reaction also forms \(\ce{(MnCl2)}\) and water.

In one batch, a technician adds 87.0 g of \(\ce{MnO2}\) to 400.0 mL of 5.00 M \(\ce{HCl}\). The chlorine produced is collected as a dry gas in a cylinder at 25\(^{\circ}\)C and 100 kPa.

If the process operates at 80% efficiency (overall yield), what volume of chlorine gas will actually be obtained?   (5 marks)

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\(9.92\ \text{L}\)

Show Worked Solution
  • The balanced chemical equation for the reaction is given by:
  •    \(\ce{MnO2 (s)+ 4HCl (aq) -> MnCl2 (aq) + Cl2 (g) + 2H2O (l)}\)
  •    \(n\ce{(MnO2)} = \dfrac{m}{MM} = \dfrac{87.0}{86.94} = 1.00\ \text{mol}\)
  •    \(n\ce{(HCl)} = c \times V = 5.00 \times 0.400 = 2\ \text{mol}\)
     
  • By the ratios in the chemical equation, 1 mole of manganese dioxide reacts with 4 moles of hydrochloric acid. As there are only 2 moles of hydrochloric acid present in the reaction, \(\ce{HCl}\) is the limiting reagent.
  •    \(n\ce{(Cl2)} = \dfrac{n\ce{(HCl)}}{4} = \dfrac{2}{4} = 0.5\ \text{mol}\)
     
  • As the process is 80% efficient, 0.4 mol of \(\ce{Cl2}\) are produced.
  •    \(V = n \times V_m = 0.4 \times 24.79 = 9.92\ \text{L (3 sig fig)}\)

CHEMISTRY, M2 EQ-Bank 6 MC

In a chemical reaction, the limiting reagent is the reactant that

  1. is present in the greatest mass
  2. is present in the smallest mass
  3. is completely consumed first, stopping the reaction
  4. determines the colour change at the endpoint
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\(C\)

Show Worked Solution
  • The limiting reagent is the reactant that is used up first in a chemical reaction, so it limits how much product can be formed.
  • Once the limiting reagent is consumed, the reaction stops even if the other reactants are still present in excess.

\(\Rightarrow C\)

CHEMISTRY, M2 EQ-Bank 7

Zinc reacts with hydrochloric acid according to the equation:

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

  1. What mass of hydrochloric acid will react with 6.54 g of zinc?   (3 marks)
  1. If 5.00 g of hydrochloric acid is reacted with 6.54 g of zinc, which is the limiting reagent?   (1 mark)
  1. What mass of hydrogen gas will be produced using the quantities in part b?   (2 marks)
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a.    \(7.29\ \text{g}\)

b.    Hydrochloric acid is the limiting reagent.

c.    \(0.138\ \text{g}\)

Show Worked Solution

a.    \(n\ce{(Zn)} = \dfrac{m}{MM} = \dfrac{6.54}{65.38} = 0.100\ \text{mol}\).

As the molar ratio is \(1\ \text{mol}\ \ce{Zn}: 2\ \text{mol}\ \ce{HCl}\),

\(n\ce{(HCl)} = 2 \times 0.100 = 0.200\ \text{mol}\).

\(m\ce{(HCl)} = 0.200 \times (1.008 + 35.45) = 7.29\ \text{g}\).
 

b.    5.00 g of \(\ce{HCl}\) is present while 7.29 g is required

\(\Rightarrow\) Hydrochloric acid is the limiting reagent.
 

c.    Since hydrochloric acid is the limiting reagent,

\(n\ce{(HCl)} = \dfrac{5.00}{36.458} = 0.13714\ \text{mol}\).

Molar ratio between \(\ce{HCl}\) and \(\ce{H2}\) is \(2\ \text{mol}\ \ce{HCl}: 1\ \text{mol}\ \ce{H2}\),

\(n\ce{(H2)} = \dfrac{1}{2} \times 0.13714 = 0.0686\ \text{mol}\).

\(m\ce{(H2)} = 0.0686 \times 2.016 = 0.138\ \text{g}\).

CHEMISTRY, M2 2006 HSC 18

A student studying the mass change that occurs during fermentation added glucose, water and yeast to a flask and stoppered the flask with some cotton wool.

The student measured the mass of the flask daily for seven days. The table shows the data collected.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \ \textit{Day}\ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \textit{Mass}\ \text{(g)}\ \ \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 381.05\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 376.96\\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 373.42\\
\hline
\rule{0pt}{2.5ex} 4 \rule[-1ex]{0pt}{0pt} & 370.44\\
\hline
\rule{0pt}{2.5ex} 5 \rule[-1ex]{0pt}{0pt} & 370.42\\
\hline
\rule{0pt}{2.5ex} 6 \rule[-1ex]{0pt}{0pt} & 370.40\\
\hline
\rule{0pt}{2.5ex} 7 \rule[-1ex]{0pt}{0pt} & 370.39\\
\hline
\end{array}

  1. Calculate the moles of \(\ce{CO2}\) released between days 1 and 7.  (1 mark)

  2. Calculate the mass of glucose that underwent fermentation between days 1 and 7. Include a balanced chemical equation in your answer.  (3 marks)

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a.    59.37 moles

b.    5395.92 g

Show Worked Solution

a.    \(\ce{MM (CO2) = 12.01 + 2 \times 16.0 = 44.01 g mol^{-1}}\)

\(\ce{Mass CO2 released = 2613.08 g}\)

\[\ce{n(CO2 released) = \frac{2613.08}{44.01} = 59.37 moles}\]  

b.    \(\ce{C6H12O6 \rightarrow 2C2H6O + 2CO2}\)

  • The moles of \(\ce{CO2}\) released and the reaction’s molar ratio can be used to calculate the mass of glucose that underwent fermentation.
  • Molar ratio between glucose and carbon dioxide = \(1:2\)
  •    \(\ce{n(C6H12O6) = \dfrac{1}{2} \times 59.37 = 29.685\ \text{mol}} \)
  •    \(\ce{MM(C6H12O6) = 180.56\ \text{g mol}^{-1}} \)
  •    \(\ce{m(C6H12O6) = 29.685 \times 180.56 = 5395.92\ \text{g}} \)

CHEMISTRY, M2 2009 HSC 22

The nitrogen content of bread was determined using the following procedure:

    • A sample of bread weighing 2.80 g was analysed.
    • The nitrogen in the sample was converted into ammonia.
    • The ammonia was collected in 50.0 mL of 0.125 mol L\(^{-1}\) hydrochloric acid. All of the ammonia was neutralised, leaving an excess of hydrochloric acid.
    • The excess hydrochloric acid was titrated with 23.30 mL of 0.116 mol L\(^{-1}\) sodium hydroxide solution.
  1. Write balanced equations for the TWO reactions involving hydrochloric acid.   (2 marks)

  2. Calculate the moles of excess hydrochloric acid.   (1 mark)

  3. Calculate the moles of ammonia.   (2 marks)

  4. Calculate the percentage by mass of nitrogen in the bread.   (2 marks)

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a.    \(\ce{NH3(g) + HCl(aq) \rightarrow NH4+(aq) + Cl-(aq)}\)

\(\ce{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H2O(l)}\)

b.   \(2.70 \times 10^{-3}\)

c.   \(3.55 \times 10^{-3}\)

d.   \(1.78\%\)

Show Worked Solution

a.    \(\ce{NH3(g) + HCl(aq) \rightarrow NH4+(aq) + Cl-(aq)}\)

\(\ce{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H2O(l)}\)
 

b.    \(\ce{n(NaOH excess) = c \times V = 0.116 \times 0.02330 = 2.70 \times 10^{-3} moles}\)
 

c.    \(\ce{n(HCl original) = c \times V = 0.125 \times 0.0500 = 6.25 \times 10^{-3} moles}\)

\(\ce{n(HCl used) = 6.25 \times 10^{-3}-2.70 \times 10^{-3} = 3.55 \times 10^{-3} moles}\)

\(\ce{n(NH3) = n(HCl used) = 3.55 \times 10^{-3} moles}\)
 

d.    \(\ce{n(N) = n(NH3) = 3.55 \times 10^{-3} moles}\)

\(\ce{Mass N = 3.55 \times 10^{-3} \times 14.01 = 0.0497 g}\)

\[\ce{\% N (by mass) = \frac{0.0497}{2.80} \times 100\% = 1.78\%}\]

CHEMISTRY, M2 2013 HSC 4 MC

Butan-1-ol burns in oxygen according to the following equation.

\(\ce{C4H9OH(l)} + \ce{6O2(g)} \rightarrow \ce{4CO2(g)} + \ce{5H2O(l)} \)

How many moles of carbon dioxide would form if two moles of butan-1-ol were burnt in excess oxygen?

  1. 2
  2. 4
  3. 8
  4. 10
Show Answers Only

\(C\)

Show Worked Solution

\(n(\ce{CO2(g)})=4 \times n(\ce{C4H9OH(l)})\)

\(\Rightarrow C\)