considering the combustion reaction of propane again, what mass of propane (c₃h₈) is necessary to react with…

Explanation:

Step1: Calculate molar masses

The molar mass of $O_2$ is $M_{O_2}=2\times16\ g/mol = 32\ g/mol$. The molar mass of $C_3H_8$ is $M_{C_3H_8}=3\times12 + 8\times1\ g/mol=44\ g/mol$.

Step2: Calculate moles of oxygen

The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}$, where $m_{O_2} = 4.53\ g$. So $n_{O_2}=\frac{4.53\ g}{32\ g/mol}=0.1415625\ mol$.

Step3: Use mole - ratio from the balanced equation

From the balanced equation $C_3H_8 + 5O_2\rightarrow3CO_2+4H_2O$, the mole - ratio of $C_3H_8$ to $O_2$ is $\frac{n_{C_3H_8}}{n_{O_2}}=\frac{1}{5}$. So $n_{C_3H_8}=\frac{1}{5}n_{O_2}$. Substituting $n_{O_2}=0.1415625\ mol$, we get $n_{C_3H_8}=\frac{0.1415625\ mol}{5}=0.0283125\ mol$.

Step4: Calculate mass of propane

The mass of $C_3H_8$, $m_{C_3H_8}=n_{C_3H_8}\times M_{C_3H_8}$. Substituting $n_{C_3H_8}=0.0283125\ mol$ and $M_{C_3H_8}=44\ g/mol$, we get $m_{C_3H_8}=0.0283125\ mol\times44\ g/mol = 1.25\ g$.

Answer:

$1.25\ g\ C_3H_8$