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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)

Show Answers Only

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

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

\(\Rightarrow C\)