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\ (1\ mol\ CO_(2))/(44.01g\ CO_(2)) \\\\=1.633719609`

3) multiply by molar ratio

`=1.633719609\ mol\ CO_(2)\ x\ (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\ (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