Question

2c₄h₁₀ + 13o₂ → 8co₂ + 10h₂o
what mass of c₄h₁₀ is used when
60.0 g o₂ reacts?
? g c₄h₁₀

Explanation:

Step1: Calculate moles of \( O_2 \)

Molar mass of \( O_2 \) is \( 32.00 \, g/mol \). Moles of \( O_2 = \frac{mass}{molar\ mass} = \frac{60.0\ g}{32.00\ g/mol} = 1.875\ mol \).

Step2: Use stoichiometric ratio

From the equation, \( 13\ mol\ O_2 \) reacts with \( 2\ mol\ C_4H_{10} \). Moles of \( C_4H_{10} = \frac{2}{13} \times moles\ of\ O_2 = \frac{2}{13} \times 1.875\ mol \approx 0.2904\ mol \).

Step3: Calculate mass of \( C_4H_{10} \)

Molar mass of \( C_4H_{10} \) is \( (4\times12.01 + 10\times1.008)\ g/mol = 58.12\ g/mol \). Mass = moles × molar mass = \( 0.2904\ mol \times 58.12\ g/mol \approx 16.9\ g \).

Answer:

\( \approx 16.9 \)