Question

The amount of carbon dioxide evolved upon complete combustion of \(116\text{ g}\) of \(n\)-butane is
(Given: atomic mass in amu \(\text{H} = 1\), \(\text{C} = 12\) and \(\text{O} = 16\))

• A. \(362\text{ g}\)
• B. \(352\text{ g}\)
• C. \(322\text{ g}\)
• D. \(176\text{ g}\)

Hint:

Always use standard mole conversions to simplify stoichiometric calculations: \[ \text{Mass of product} = \left(\frac{\text{Mass of Reactant}}{\text{Molar Mass of Reactant}}\right) \times (\text{Mole Ratio}) \times (\text{Molar Mass of Product}) \] Here: \(\frac{116}{58} \times 4 \times 44 = 2 \times 4 \times 44 = 352\text{ g}\).

Answer 1

The Correct Option is B

Step 1: Write the combustion equation.
n-butane is \(C_4H_{10}\). Its balanced combustion is \[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \] So one mole of butane gives four moles of \(CO_2\).
Step 2: Find the molar mass of butane.
\(C_4H_{10} = 4(12) + 10(1) = 48 + 10 = 58\ g\,mol^{-1}\).
Step 3: Find the moles of butane.
\[ n = \frac{116}{58} = 2\ \text{mol} \]
Step 4: Find the moles of CO2.
Each mole of butane gives 4 moles of \(CO_2\), so \[ 2 \times 4 = 8\ \text{mol of } CO_2 \]
Step 5: Find the molar mass of CO2.
\(CO_2 = 12 + 2(16) = 44\ g\,mol^{-1}\).
Step 6: Find the mass of CO2.
\[ \text{mass} = 8 \times 44 = 352\ g \] This is option 2.
\[ \boxed{352\ \text{g of } CO_2} \]