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When ethanol \(\ce{(C2H5OH)}\) reacts with concentrated sulfuric acid in the presence of oxygen, it undergoes a charring reaction to form solid carbon, carbon dioxide, sulfur dioxide and water:
\(\ce{C2H5OH (l) + 2H2SO4 (aq) + O2 (g) -> C (s) + CO2 (g) + 5H2O (l) + 2SO2 (g)}\)
What mass of carbon could be produced from the reaction of 9.2 g of ethanol with excess sulfuric acid and oxygen?
\(C\)
\(MM\ce{(C2H5OH)} = 2(12.01) + 6(1.008) + 16.00 = 46.068\ \text{g mol}^{-1}\).
\(n\ce{(C2H5OH)} = \dfrac{9.2}{46.068} = 0.200\ \text{mol}\).
\(\Rightarrow C\)
A 1.50 kg sample of natural gas contains 600.0 g of methane \(\ce{(CH4)}\) and the remainder is ethane \(\ce{(C2H6)}\).
What mass of this sample is due to carbon?
\(B\)
\(MM\ce{(CH4)} = 12.01 + 4(1.008) = 16.042\ \text{g mol}^{-1}\).
\(n\ce{(CH4)} = \dfrac{m}{MM} = \dfrac{600}{16.042} = 37.402\ \text{mol}\)
\(\Rightarrow B\)
Zinc reacts with hydrochloric acid according to the equation:
\(\ce{Zn (s) + 2HCl (aq) -> ZnCl2 (aq) + H2 (g)}\)
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a. \(7.29\ \text{g}\)
b. Hydrochloric acid is the limiting reagent.
c. \(0.138\ \text{g}\)
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 \(2\ \text{mol}\ \ce{HCl}: 1\ \text{mol}\ \ce{H2}\),
\(n\ce{(H2)} = \dfrac{1}{2} \times 0.13714 = 0.0686\ \text{mol}\).
\(m\ce{(H2)} = 0.0686 \times 2.016 = 0.138\ \text{g}\).
Aluminium oxide can be formed when aluminium metal reacts with oxygen gas according to the equation:
\(\ce{4Al(s) + 3O2(g) \rightarrow 2Al2O3(s)}\)
In an experiment, 2.70 g of aluminium is heated strongly in oxygen.
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a. \(0.10\ \text{mol}\)
b. \( 2.40\ \text{g}\)
c. \(5.10\ \text{g}\)
a. \(n\ce{(Al)} = \dfrac{m}{MM} = \dfrac{2.70}{26.98} = 0.10\ \text{mol}\)
b. The molar ratio of aluminium to oxygen is \(4:3\).
Hence, \(n\ce{(O)} = \dfrac{3}{4} \times n\ce{(Al)} = \dfrac{3}{4} \times 0.10 = 0.075\).
\(m\ce{(O2)} = n \times MM = 0.075 \times 32.00 = 2.40\ \text{g}\).
c. \(n\ce{(Al2O3)} = \dfrac{2}{4} \times n\ce{(Al)} = \dfrac{1}{2} \times 0.10 = 0.05\ \text{mol}\).
\(m\ce{(Al2O3)} = n \times MM = 0.05 \times (2(26.98) + 3(16.00)) = 5.10\ \text{g}\).
A gardener spreads a 2.00 kilogram bag of ammonium sulfate fertilizer, \(\ce{(NH4)_2SO4}\), over a lawn.
What is the number of moles of \(\ce{(NH4)_2SO4}\) in the bag?
\(C\)
\(\Rightarrow C\)
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.
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:
\(\ce{N(O2) = n \times N_A = 1.00 \, mol \times 6.022 \times 10^{23} \, mol^{-1} = 6.022 \times 10^{23} \, molecules}\)
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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\(226.8\ \text{mol}\)
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?
\(C\)
\(\Rightarrow C\)
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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\(\text{0.062 grams}\)
\(\ce{n(C) = \dfrac{0.157}{12.0} = 0.0131\ \text{mol}} \)
\(\ce{n(\text{Taxol}) = \dfrac{0.0131}{47}\ \text{mol}} \)
\(\ce{n(O) = 14 \times n(\text{Taxol}) = 14 \times \dfrac{0.0131}{47} = 0.00390\ \text{mol}} \)
\( \therefore \ce{m(O) = 0.00390 \times 16.0 = 0.062\ \text{g (3 d.p.)}}\)
Calculate the mass of solid sodium hydrogen carbonate required to make 250 mL of 0.12 mol L\(^{-1}\) solution. (2 marks)
\(2.52\ \text{g} \)
\(\ce{n(NaHCO3\ \text{in solution})\ = 0.12 \times \dfrac{250}{1000} = 0.03\ \text{mol}}\)
\(\ce{MM(NaHCO3) = 22.99 + 1.008 + 12.01 + 3 \times 16.00 = 84.0\ \text{g mol}^{-1}} \)
\(\ce{m(NaHCO3) = n \times MM = 0.03 \times 84.0 = 2.52\ \text{g}} \)
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\(^{-1}\) ?
\(B\)
\(\Rightarrow B\)
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}
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a. 59.37 moles
b. 5395.92 g
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 nitrogen content of bread was determined using the following procedure:
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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. \(1.78\%\)
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. \(\ce{n(N) = n(NH3) = 3.55 \times 10^{-3} moles}\)
\(\ce{Mass N = 3.55 \times 10^{-3} \times 14.01 = 0.0497 g}\)
\[\ce{\% N (by mass) = \frac{0.0497}{2.80} \times 100\% = 1.78\%}\]
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?
\(B\)
| \(MM(\ce{C6H12O6})\) | \(= 6(12.01)+12(1.008)+6(16)\) | |
| \(=180.156\ \text{gmol}^{-1}\) |
\(n(\ce{C6H12O6})=\dfrac{6.50}{180.156}=0.036\ \text{mol}\)
\(n(\ce{CO2})=0.072\ \text{mol}\)
\(m(\ce{CO2})=0.072 \times 44.01=3.18\ \text{g}\)
\(\Rightarrow B\)
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 ?
\(\ce{C6H12O6 \rightarrow 2CO2 + 2C2H5OH}\)
\(B\)
\(n(\ce{CO2})= \dfrac{5.68}{44.01} =0.129\ \text{mol} = n(\ce{C2H5OH})\)
\(m(\ce{C2H5OH})=0.129 \times 46.068 = 5.95\ \text{g}\)
\(\Rightarrow B\)