Answer:
[tex]\large \boxed{\text{748 g}}[/tex]
Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Máµ£: Â Â 44.10 Â Â Â Â Â Â Â Â 44.01
    C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
m/g: Â 250.
(a) Moles of C₃H₈
[tex]\text{Moles of C$_{3}$H}_{8} = \text{250. g C$_{3}$H}_{8}\times \dfrac{\text{1 mol C$_{3}$H}_{8}}{\text{44.10 g C$_{3}$H}_{8}}= \text{5.669 mol C$_{3}$H}_{8}[/tex]
(b) Moles of COâ‚‚
[tex]\text{Moles of CO}_{2} = \text{5.669 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} = \text{17.01 mol CO}_{2}[/tex]
(c) Mass of COâ‚‚
[tex]\text{Mass of CO}_{2} =\text{17.01 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{748 g CO}_{2}\\\\\text{The reaction can produce $\large \boxed{\textbf{748 g}}$ of CO}_{2}[/tex]
