c₅h₁₂ + 8o₂ → 5co₂ + 6h₂o\nhow many moles of oxygen would be required to completely combust 521 grams of…

Explanation:

Step1: Calculate moles of pentane

The molar mass of pentane ($C_5H_{12}$) is $M=(5\times12 + 12\times1)\text{ g/mol}=72\text{ g/mol}$. The number of moles of pentane $n_{C_5H_{12}}=\frac{m}{M}$, where $m = 521\text{ g}$. So $n_{C_5H_{12}}=\frac{521\text{ g}}{72\text{ g/mol}}\approx7.24\text{ mol}$.

Step2: Use mole - ratio from balanced equation

From the balanced equation $C_5H_{12}+8O_2\rightarrow5CO_2 + 6H_2O$, the mole - ratio of $C_5H_{12}$ to $O_2$ is $1:8$. So if $n_{C_5H_{12}} = 7.24\text{ mol}$, then the number of moles of $O_2$, $n_{O_2}=8\times n_{C_5H_{12}}$. $n_{O_2}=8\times7.24\text{ mol}=57.92\text{ mol}$

Answer:

$57.92$ mol