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Ammonia \(\ce{(NH3)}\) is synthesized from nitrogen \(\ce{(N2)}\) and hydrogen \(\ce{(H2)}\) according to the following equation:
\(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\)
If 1.50 moles of nitrogen gas react completely with hydrogen gas, calculate the mass of ammonia produced. (2 marks)
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\(\ce{m(NH3) = 51.0\ \text{g}}\)
Calculate the number of moles of ammonia \(\ce{(NH3)}\) produced:
\(\ce{n(NH3) = 2 \times n(N2) = 2 \times 1.50 \, mol = 3.00 \, mol}\)
Calculate the mass of ammonia produced:
\(\ce{m(NH3) = n \times MM = 3.00 \, mol \times 17.03\ \text{g mol}^{-1} = 51.09\ \text{g}}\)
→ The mass of ammonia produced is 51.0 grams.
Calculate the mass of solid sodium hydrogen carbonate required to make 250 mL of 0.12 mol L\(^{-1}\) solution. (2 marks)
\(2.52\ \text{g} \)
\(\ce{n(NaHCO3\ \text{in solution})\ = 0.12 \times \dfrac{250}{1000} = 0.03\ \text{mol}}\)
\(\ce{MM(NaHCO3) = 22.99 + 1.008 + 12.01 + 3 \times 16.00 = 84.0\ \text{g mol}^{-1}} \)
\(\ce{m(NaHCO3) = n \times MM = 0.03 \times 84.0 = 2.52\ \text{g}} \)
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 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 = \(1:2\)
\(\ce{n(C6H12O6) = \dfrac{1}{2} \times 59.37 = 29.685\ \text{mol}} \)
\(\ce{MM(C6H12O6) = 180.56\ \text{g mol}^{-1}} \)
\(\ce{m(C6H12O6) = 29.685 \times 180.56 = 5395.92\ \text{g}} \)
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?
\(B\)
| \(MM(\ce{C6H12O6})\) | \(= 6(12.01)+12(1.008)+6(16)\) | |
| \(=180.156\ \text{gmol}^{-1}\) |
\(n(\ce{C6H12O6})=\dfrac{6.50}{180.156}=0.036\ \text{mol}\)
\(n(\ce{CO2})=0.072\ \text{mol}\)
\(m(\ce{CO2})=0.072 \times 44.01=3.18\ \text{g}\)
\(\Rightarrow B\)