Question

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

Answer 1

`\boxed{28.81}`

Step-by-step 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) $} * \frac{\text{1 mol C$_(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) $} * \frac{\text{8 mol CO}_(2)}{\text{2 mol C$_(4)$H$_(10)$}} = \text{0.6546 mol CO}_(2)`

3. Mass of CO₂

`\text{Mass of CO}_(2) = \text{0.6546 mol CO}_(2) * \frac{\text{44.01 g CO}_(2)}{\text{1 mol CO}_(2)} = \textbf{28.81 g CO}_(2)\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_(2)}$}`