Question

For the complete combustion of 47 g of gasoline (octane, C8H18) , the mass of oxygen consumed is

Answer 1

Answer: d) 164.9 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$     $\text{Moles of octane}=\frac{47g}{114g/mol}=0.412moles$

The balanced chemical reaction is:

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)$  

According to stoichiometry :

2 moles of $C_8H_{18}$ require = 25  moles of $O_2$

Thus 0.412 moles of $C_8H_{18}$ will require=$\frac{25}{2}\times 0.412=5.15moles$  of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=5.15moles\times 32g/mol=164.9g$

Thus 164.9 g of oxygen is consumed.