Answer:
[tex]\boxed{28.81}[/tex]
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
[tex]\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}$}[/tex]
2. Moles of CO₂
The molar ratio is 8 mol CO₂:2 mol C₄H₁₀
[tex]\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}[/tex]
3. Mass of CO₂
[tex]\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 $\boxed{\textbf{28.81 g CO}_{2}}$}[/tex]
