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

Concept: Stoichiometry allows us to calculate mass relationships in a balanced chemical equation. Complete combustion of any alkane hydrocarbon produces carbon dioxide (\(\text{CO}_2\)) gas and water vapor (\(\text{H}_2\text{O}\)) as the only products.

Step 1: Write and balance the combustion equation for \(n\)-butane
The molecular formula for \(n\)-butane is \(\text{C}_4\text{H}_{10}\). Let's construct its balanced combustion equation with gaseous oxygen (\(\text{O}_2\)): \[ \text{C}_4\text{H}_{10} + \frac{13}{2}\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O} \] Multiplying through by 2 to clear the fraction yields integer stoichiometric coefficients: \[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]

Step 2: Calculate the molar masses of the relevant substances
Using the given atomic masses, we compute:

• Molar mass of \(n\)-butane (\(\text{C}_4\text{H}_{10}\)): \[ \text{Molar Mass} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58\text{ g/mol} \]

• Molar mass of Carbon Dioxide (\(\text{CO}_2\)): \[ \text{Molar Mass} = (1 \times 12) + (2 \times 16) = 12 + 32 = 44\text{ g/mol} \]

Step 3: Determine the total number of moles of butane reacted
We are given an initial mass of \(116\text{ g}\) of \(n\)-butane. \[ \text{Number of moles of }\text{C}_4\text{H}_{10} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{116\text{ g}}{58\text{ g/mol}} = 2\text{ moles} \]

Step 4: Use molar ratios to find the mass of evolved \(\text{CO}_2\)
From our balanced chemical equation, 1 mole of \(\text{C}_4\text{H}_{10}\) completely yields 4 moles of \(\text{CO}_2\). Therefore, the number of moles of \(\text{CO}_2\) produced by 2 moles of butane is: \[ \text{Moles of }\text{CO}_2 = 2\text{ moles of }\text{C}_4\text{H}_{10} \times 4 = 8\text{ moles} \] Now, convert the moles of carbon dioxide into total mass in grams: \[ \text{Mass of }\text{CO}_2 = \text{Moles} \times \text{Molar Mass} \] \[ \text{Mass of }\text{CO}_2 = 8\text{ moles} \times 44\text{ g/mol} = 352\text{ g} \] Thus, the total amount of carbon dioxide evolved is equal to \(352\text{ g}\).