Question

Calculate the mass of butane needed to produce 71.9 g of carbon dioxide.

Answer 1

1) balanced chemical equation

$2\ C_{4}H_{10}_{(g)}\ +\ 13\ O_{2}_{(g)}\ ->\ 10\ H_{2}O_{(g)}\ +\ 8\ CO_{2}_{(g)}$

2) convert mass of CO₂ to moles

$=71.9g\ CO_{2}\ x\ \frac{1\ mol\ CO_{2}}{44.01g\ CO_{2}} \\\\=1.633719609$

3) multiply by molar ratio

$=1.633719609\ mol\ CO_{2}\ x\ \frac{2\ mol\ C_{4}H_{10}}{8\ mol\ CO_{2}}\\\\=0.4084299023$

4) convert moles of C₄H₁₀ to mass

$=0.4084299023\ mol\ C_{4}H_{10}\ x\ \frac{58.14g\ C_{4}H_{10}}{1\ mol\ C_{4}H_{10}}\\\\=23.74611452$

= 23.7 grams of C₄H₁₀ is needed to produce 71.9 grams of CO₂