Question

Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2

Answer 1

Answer:

$\boxed{28.81}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

$\text{Moles of C _{4} H _{10} } = \text{ 9.511 g C _{4} H _{10} } \times \dfrac{\text{1 mol C _{4} H _{10} }}{\text{ 58.12 g C _{4} H _{10} }} = \text{0.1636 mol C _{4} H _{10} }$

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

$\text{Moles of CO}_{2} =\text{0.1636 mol C _{4} H _{10} } \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C _{4} H _{10} }} = \text{0.6546 mol CO}_{2}$

3. Mass of CO₂

$\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form \boxed{\textbf{28.81 g CO}_{2}} }$