Question

consider the combustion of propane with oxygen: 2 c₃h₈ + 7 o₂ --> 6 co₂ + 8 h₂o. if you have 5.0 mol of propane and 10.0 mol of oxygen, which reactant is the limiting reactant? oxygen carbon dioxide propane water

Explanation:

Step1: Determine mole - ratio from equation

The balanced equation is $2C_3H_8 + 7O_2
ightarrow6CO_2 + 8H_2O$. The mole - ratio of $C_3H_8$ to $O_2$ is $\frac{n_{C_3H_8}}{n_{O_2}}=\frac{2}{7}$.

Step2: Calculate moles of $O_2$ needed for given $C_3H_8$

Given $n_{C_3H_8}=5.0$ mol. Using the mole - ratio, the moles of $O_2$ required for complete reaction of $C_3H_8$ is $n_{O_2\ required}=\frac{7}{2}\times n_{C_3H_8}=\frac{7}{2}\times5.0 = 17.5$ mol.

Step3: Compare available and required moles of $O_2$

We have $n_{O_2\ available}=10.0$ mol. Since $10.0\ mol<17.5\ mol$, oxygen is the limiting reactant.

Answer:

A. oxygen