Answer:
[tex]\boxed{\text{190 g}}[/tex]
Explanation:
We know we will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 44.01
2C₆H₁₀ + 17O₂ ⟶ 12CO₂ + 10H₂O
n/mol: 6
1. Use the molar ratio of CO₂:O₂ to calculate the moles of CO₂.
[tex]\text{Moles of CO$_{2}$ = 6 mol O$_{2}$} \times \dfrac{\text{12 mol CO$_{2}$}}{\text{17 mol O$_{2}$}} =\text{4.24 mol CO$_{2}$}[/tex]
2. Use the molar mass of CO₂ to calculate the mass of CO₂.
[tex]\text{Mass of CO$_{2}$ = 4.24 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \text{190 g CO$_{2}$}\\\\\text{The reaction will produce }\boxed{\textbf{190 g}}\text{of CO}_{2}[/tex]
