Butane reacts with oxygen to produce carbon dioxide and water following the equation given below:
$
C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)
$
If 174.0 kg of butane is mixed with 320.0 kg of $ O_2 $, the volume of water formed in litres is _____. (Nearest integer)
[Given: (a) Molar masses: C = 12, H = 1, O = 16 g $ mol^{-1} $; (b) Density of water = 1 g $ mL^{-1} $]
$ C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l) $ If 174.0 kg of butane is mixed with 320.0 kg of $ O_2 $, the volume of water formed in litres is _____. (Nearest integer)
[Given: (a) Molar masses: C = 12, H = 1, O = 16 g $ mol^{-1} $; (b) Density of water = 1 g $ mL^{-1} $]
Show Hint
Correct Answer: 138
Approach Solution - 1
Moles of \( C_4H_{10} = \frac{174 \times 10^3 \, g}{58 \, g/mol} = 3000 \, mol \) Moles of \( O_2 = \frac{320 \times 10^3 \, g}{32 \, g/mol} = 10000 \, mol \)
Step 2: Identify the limiting reactant.
From the balanced equation, 1 mol \( C_4H_{10} \) reacts with 6.5 mol \( O_2 \). 3000 mol \( C_4H_{10} \) requires \( 3000 \times 6.5 = 19500 \, mol \, O_2 \). Since we have only 10000 mol \( O_2 \), oxygen is the limiting reactant.
Step 3: Calculate moles of water formed.
From the stoichiometry, 6.5 mol \( O_2 \) produces 5 mol \( H_2O \). 10000 mol \( O_2 \) produces \( \frac{5}{6.5} \times 10000 = \frac{10}{13} \times 10000 \approx 7692.3 \, mol \, H_2O \).
Step 4: Calculate the volume of water formed.
Mass of \( H_2O = 7692.3 \, mol \times 18 \, g/mol \approx 138461.4 \, g \). Volume of \( H_2O = \frac{138461.4 \, g}{1 \, g/mL} = 138461.4 \, mL = 138.4614 \, L \).
Nearest integer = 138 L.
Approach Solution -2
Molar masses: \(\mathrm{M(C_4H_{10}) = 58\ g\ mol^{-1}}\), \(\mathrm{M(O_2) = 32\ g\ mol^{-1}}\).
Moles of butane: \(\displaystyle n_{\mathrm{C_4H_{10}}}=\frac{174000}{58}=3000\ \text{mol}\).
Moles of oxygen: \(\displaystyle n_{\mathrm{O_2}}=\frac{320000}{32}=10000\ \text{mol}\).
Required O2 for 3000 mol butane: \(3000\times 6.5=19500\) mol > available 10000 mol ⇒ O2 is limiting.
Moles of butane that actually react: \(\displaystyle n_{\text{react}}=\frac{10000}{6.5}=1538.46\ \text{mol}\).
From stoichiometry, water formed: \(5\) mol H2O per mol butane ⇒
\(\displaystyle n_{\mathrm{H_2O}}=5\times 1538.46=7692.31\ \text{mol}\).
Mass of water: \(\displaystyle m=7692.31\times 18=138461.6\ \text{g}=138.46\ \text{kg}\).
Density \(=1\ \text{g mL}^{-1}\) ⇒ volume \(=138461.6\ \text{mL}=138.46\ \text{L}\).
Final Answer:
\[ \boxed{138\ \text{L}} \]
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