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In an experiment, 5.85 g of ethanol was ignited with 14.2 g of oxygen.
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i. \(\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }\)
i. \(\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }\)
ii. \(\ce{MM(C2H5OH) = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\ \text{g mol}^{-1}}\)
\(\ce{n(C2H5OH) = \dfrac{5.85}{46.0} = 0.1272\ \text{mol}} \)
\(\ce{n(O2) = \dfrac{14.2}{32.0} = 0.444\ \text{mol}}\)
\(\text{Reaction ratio}\ \ \ce{C2H5OH : O2 = 1:3}\)
\(\Rightarrow \ce{n(O2)_{\text{required}} = 3 \times 0.1272 = 0.382\ \text{mol}\ \ (\ce{O2}\ \text{excess}) }\)
\(\ce{O2_{\text{(excess)}} = 0.444-0.382 = 0.062\ \text{mol}}\)
When concentrated sulfuric acid is added to dry sucrose, \(\ce{C12H22O11}\), a black residue of pure carbon is produced.
An equation for the reaction is
\(\ce{2C12H22O11(s)+ 2H2SO4(aq) + O2(g)\rightarrow 22C(s) + 2CO2(g) + 24H2O(g) + 2SO2(g)}\)
\(\text{Molar mass}\ \ce{(C12H22O11) = 342.0 \text{ g mol}^{–1}}\)
Calculate the mass of carbon residue that could be produced by the reaction of 50.0 g of sucrose with excess concentrated sulfuric acid. (2 marks)
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\(\text{19.3 grams}\)
\(\ce{n(C12H22O11)_{\text{react}} = \dfrac{50.0}{342.0} = 0.146\ \text{mol}} \)
\(\ce{n(C)_{\text{produced}} = 11 \times n(C12H22O11) = 11 \times 0.146 = 1.61\ \text{mol}} \)
\(\ce{m(C)_{\text{produced}} = 1.61 \times 12.0 = 19.3\ \text{g}} \)
Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation.
\(\ce{P2O5}(s) + \ce{3H2O}(l) \rightarrow \ce{2H3PO4}(aq)\)
Phosphoric acid reacts with sodium hydroxide according to the following equation.
\(\ce{H3PO4}(aq) + \ce{3NaOH}(aq) \rightarrow \ce{Na3PO4}(aq) + \ce{3H2O}(l)\)
A student reacted 1.42 g of phosphorus pentoxide with excess water.
What volume of 0.30 mol L\(^{-1}\) sodium hydroxide would be required to neutralise all the phosphoric acid produced?
\(C\)
| \(MM(\ce{P2O5})\) | \(=2(30.97)+5(16)\) | |
| \(=141.94\ \text{g\mol}\) |
\(n(\ce{P2O5})=\dfrac{1.42}{141.94}=0.01 \ \text{mol}\)
\(n(\ce{H3PO4}) = 2 \times 0.01 = 0.02\ \text{mol}\)
\(n(\ce{NaOH})\) for neutralisation \(= 3 \times 0.02 = 0.06\ \text{mol}\)
| \(c\) | \(=\dfrac{n}{V}\) | |
| \(V\) | \(=\dfrac{n}{c}\) | |
| \(=\dfrac{0.06}{0.3}\) | ||
| \(=0.2\ \text{L}\) |
\( \Rightarrow C\)
When charcoal reacts in the presence of oxygen, carbon monoxide and carbon dioxide are produced according to the following chemical reactions.
\( \ce{C}(s) +\frac{1}{2} \ce{O}_2(g) \rightarrow \mathrm{CO}(g)\)
\(\ce{C}(s) +\ce{O2}(g) \rightarrow \ce{CO2}(g)\)
What would be the total mass of gas produced when 400 g of charcoal is reacted, assuming equal amounts are consumed in each reaction?
\(B\)
\(n(\ce{C} (s))=\dfrac{400}{12.01}=33.3\ \text{mol}\)
→ 16.65 \(\text{mol}\) of charcoal goes into each reaction. Therefore 16.65 \(\text{mol}\) of \(\ce{CO}(g)\) and \(\ce{CO2}(g)\) are produced.
\(m(\ce{CO})=16.65 \times 28.01 =466.4\ \text{g}\)
\(m(\ce{CO2})=16.65 \times 44.01 =732.8\ \text{g}\)
→ Total mass produced = 1199.2 \(\text{g}\) = 1.2 \(\text{kg}\).
\(\Rightarrow B\)
Under conditions of low oxygen levels, octane can undergo incomplete combustion according to the following chemical equation:
\( \ce{2C8H18}(l) + \ce{17O2}(g) \rightarrow \ce{6C}(s)+4 \ce{CO}(g) + \ce{6CO2}(g) + \ce{18H2O}(l)\)
Calculate the mass of soot \((\ce{C}(s))\) produced if 50 grams of octane are combusted in this way with 30 grams of oxygen and the mass of soot accounts for \(\dfrac{1}{5}\) of the total mass of the products. (2 marks)
\(16\ \text{g}\)
→ Total mass of the reactants 80 grams.
→ By the law of conservation of mass, the total mass of the products will be 80 grams.
→ Mass of soot \(=\dfrac{1}{5} \times 80 = 16\ \text{g}\)
Excess barium nitrate solution is added to 200 mL of 0.200 mol L\(^{-1}\) sodium sulfate.
What is the mass of the solid formed?
`C`
\( \ce{Ba(NO3)2 (aq) + Na2SO4 (aq) \rightarrow BaSO4 (s) + 2NaNO3 (aq)}\)
\(n(\ce{Na2SO4})=0.2 \times 0.2=0.04\ \text{mol}\)
\(n(\ce{BaSO4 (s)})=0.04\ \text{mol}\)
\(m(\ce{BaSO4 (s)})=0.04 \times 233.37=9.33\ \text{g}\)
\(\Rightarrow C\)