Question

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} $]

Hint:

In stoichiometry problems involving masses of reactants, always convert masses to moles first. Then, use the balanced chemical equation to identify the limiting reactant. The amount of product formed is determined by the limiting reactant. Finally, convert the moles of product back to the desired unit (in this case, volume using density).

Answer 1

Correct Answer: 138

Step 1: Calculate moles of reactants.
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.