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CHEMISTRY, M2 EQ-Bank 6

During a laboratory experiment, 32.0 grams of oxygen gas \(\ce{(O2)}\) is required. Calculate the number of molecules of oxygen gas needed.   (2 marks)

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\(6.022 \times 10^{23}\) molecules.

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Calculate the number of moles of oxygen gas (\(\ce{O2}\)):

  \(\ce{n(O2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{32.0\ \text{g}}{32.00\ \text{g mol}^{-1}} = 1.00\ \text{mol}}\)
 

Use Avogadro’s number to find the number of molecules:

  \(\ce{N(O2) = n \times N_A = 1.00 \, mol \times 6.022 \times 10^{23} \, mol^{-1} = 6.022 \times 10^{23} \, molecules}\)

→ Oxygen gas required = \(6.022 \times 10^{23}\) molecules.

CHEMISTRY, M2 EQ-Bank 4

A propane tank contains 10.0 kg of propane \(\ce{(C3H8)}\). How many moles of propane are in the tank?   (2 marks)

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 \(226.8\ \text{mol}\)

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Calculate the number of moles of propane (\(\ce{C3H8}\)):

  \(\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{10\,000\ \text{g}}{44.094 \, \text{g mol}^{-1}} = 226.79 \, mol}\)

→ The number of moles of propane in the tank is 226.8 mol.

CHEMISTRY, M2 2012 VCE 18 MC

2.1 g of an alkene that contains only one double bond per molecule reacted completely with 8.0 g of bromine, \(\ce{Br2}\). The molar mass of bromine, \(\ce{Br2}\), is 160 g mol\(^{–1}\).

Which one of the following is the molecular formula of the alkene?

  1. \(\ce{C5H10}\)
  2. \(\ce{C4H8}\)
  3. \(\ce{C3H6}\)
  4. \(\ce{C2H4}\)
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\(C\)

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→ Since each molecule only contains a single \(\ce{C=C}\) bond:

\(\ce{n(alkene) = n(Br2) = \dfrac{8.0}{160.0} = 0.05\ \text{mol}}\)

\(\ce{m(alkene) = \dfrac{2.1}{0.05} = 42\ \text{g mol}^{-1}} \)

\(\Rightarrow C\)

CHEMISTRY, M2 2012 VCE 15*

A sample of the anticancer drug Taxol\(^{\circledR}\), \(\ce{ C47H51NO14}\), contains 0.157 g of carbon.

Calculate the mass, in grams, of oxygen in the sample. Give your answer correct to three decimal places.   (2 marks)

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\(\text{0.062 grams}\)

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\(\ce{n(C) = \dfrac{0.157}{12.0} = 0.0131\ \text{mol}} \)

\(\ce{n(\text{Taxol}) = \dfrac{0.0131}{47}\ \text{mol}} \)

\(\ce{n(O) = 14 \times n(\text{Taxol}) = 14 \times \dfrac{0.0131}{47} = 0.00390\ \text{mol}} \)

\( \therefore \ce{m(O) = 0.00390 \times 16.0 = 0.062\ \text{g  (3 d.p.)}}\)

CHEMISTRY, M2 2004 HSC 16b

Calculate the mass of solid sodium hydrogen carbonate required to make 250 mL of 0.12 mol L\(^{-1}\) solution.  (2 marks)

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\(2.52\ \text{g} \)

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\(\ce{n(NaHCO3\ \text{in solution})\ = 0.12 \times \dfrac{250}{1000} = 0.03\ \text{mol}}\)

\(\ce{MM(NaHCO3) = 22.99 + 1.008 + 12.01 + 3 \times 16.00 = 84.0\ \text{g mol}^{-1}} \)

\(\ce{m(NaHCO3) = n \times MM = 0.03 \times 84.0 = 2.52\ \text{g}} \)

CHEMISTRY, M2 2005 HSC 3 MC

The heat of combustion of butan-1-ol \((\ce{C4H10O})\) is 2676 kJ mol\(^{-1}\).

What is the value of the heat of combustion in kJ g\(^{-1}\) ?

  1. 30.41
  2. 36.10
  3. 44.60
  4. 47.79
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\(B\)

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→ The molar mass of butan-1-ol is \(74.12\ \text{gmol}^{-1}\).

→ The heat of combustion in kJ g\(^{-1} = \dfrac{2676}{74.12}=36.10\)  kJ g\(^{-1}\).

\(\Rightarrow B\)

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

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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 2009 HSC 13 MC

In a fermentation experiment 6.50 g of glucose was completely converted to ethanol and carbon dioxide.

\(\ce{C6H12O6 \rightarrow 2C2H5OH + 2CO2}\)

What is the mass of carbon dioxide produced?

  1. 1.59 g
  2. 3.18 g
  3. 9.53 g
  4. 13.0 g
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\(B\)

Show Worked Solution
\(MM(\ce{C6H12O6})\) \(= 6(12.01)+12(1.008)+6(16)\)  
  \(=180.156\ \text{gmol}^{-1}\)  

 

\(n(\ce{C6H12O6})=\dfrac{6.50}{180.156}=0.036\ \text{mol}\)

\(n(\ce{CO2})=0.072\ \text{mol}\)

\(m(\ce{CO2})=0.072 \times 44.01=3.18\ \text{g}\)

\(\Rightarrow B\)

CHEMISTRY, M2 2010 HSC 15 MC

What mass of ethanol \(\ce{C2H5OH}\) is obtained when 5.68 g of carbon dioxide is produced during fermentation, at 25°C and 100 kPa ?

\(\ce{C6H12O6 \rightarrow 2CO2 + 2C2H5OH}\)

  1. 2.95 g
  2. 5.95 g
  3. 33.6 g
  4. 147.2 g
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\(B\)

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\(n(\ce{CO2})= \dfrac{5.68}{44.01} =0.129\ \text{mol} = n(\ce{C2H5OH})\)

\(m(\ce{C2H5OH})=0.129 \times 46.068 = 5.95\ \text{g}\)

\(\Rightarrow B\)