the chemical equation below shows the combustion of propane (c₃h₈). c₃h₈ + 5o₂ → 3co₂ + 4h₂o the molar mass…

Explanation:

Step1: Calculate moles of $C_3H_8$

Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass and $M$ is the molar - mass. Given $m = 0.025\ g$ and $M = 44.1\ g/mol$ for $C_3H_8$. $n_{C_3H_8}=\frac{0.025\ g}{44.1\ g/mol}\approx0.000567\ mol$

Step2: Determine mole ratio of $C_3H_8$ to $O_2$

From the balanced chemical equation $C_3H_8 + 5O_2\rightarrow3CO_2+4H_2O$, the mole ratio of $C_3H_8$ to $O_2$ is $1:5$. So, if $n_{C_3H_8}=0.000567\ mol$, then $n_{O_2}=5\times n_{C_3H_8}$. $n_{O_2}=5\times0.000567\ mol = 0.002835\ mol$

Step3: Calculate mass of $O_2$

Use the formula $m = n\times M$. Given $n = 0.002835\ mol$ and $M = 32.00\ g/mol$ for $O_2$. $m_{O_2}=0.002835\ mol\times32.00\ g/mol=0.09072\ g\approx0.091\ g$

Answer:

$0.091$ grams