the chemical equation below shows the combustion of methane (ch4).\nch4 + 2o2 → co2 + 2h2o\nthe molar mass…

Explanation:

Step1: Calculate moles of O₂

Use the formula $n=\frac{m}{M}$, where $n$ is moles, $m$ is mass, and $M$ is molar - mass. Given $m = 8.94\ g$ and $M = 32.00\ g/mol$ for O₂. $n_{O_2}=\frac{8.94\ g}{32.00\ g/mol}=0.279375\ mol$

Step2: Determine mole - ratio

From the balanced chemical equation $CH_4 + 2O_2\rightarrow CO_2+2H_2O$, the mole - ratio of $O_2$ to $CO_2$ is $2:1$. So, if $n_{O_2}=0.279375\ mol$, then $n_{CO_2}=\frac{1}{2}n_{O_2}$. $n_{CO_2}=\frac{1}{2}\times0.279375\ mol = 0.1396875\ mol$

Step3: Calculate mass of CO₂

Use the formula $m = n\times M$. Given $n = 0.1396875\ mol$ and $M = 44.01\ g/mol$ for $CO_2$. $m_{CO_2}=0.1396875\ mol\times44.01\ g/mol\approx6.15\ g$

Answer:

6.15 grams