Question

5. According to the equation C3H8 + 502

3CO2 + 4H20, when 02 is in excess, how many grams of
carbon dioxide are produced from the combustion of 250. g of ethane, C3H8?
produced from the combustion of 250 g of ethane

Answer 1

`\large \boxed{\text{748 g}}`

Step-by-step explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 44.10 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

m/g: 250.

(a) Moles of C₃H₈

`\text{Moles of C$_(3)$H}_(8) = \text{250. g C$_(3)$H}_(8)* \frac{\text{1 mol C$_(3)$H}_(8)}{\text{44.10 g C$_(3)$H}_(8)}= \text{5.669 mol C$_(3)$H}_(8)`

(b) Moles of CO₂

`\text{Moles of CO}_(2) = \text{5.669 mol C$_(3)$H}_(8) * \frac{\text{3 mol CO}_(2)}{\text{1 mol C$_(3)$H}_(8)} = \text{17.01 mol CO}_(2)`

(c) Mass of CO₂

`\text{Mass of CO}_(2) =\text{17.01 mol CO}_(2) * \frac{\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)`