methane (ch₄, 16.05 g/mol) reacts with oxygen to form carbon dioxide (co₂, 44.01 g/mol) and water (h₂o…

Explanation:

Step1: Write the balanced chemical equation

$CH_4 + 2O_2\rightarrow CO_2+2H_2O$

Step2: Calculate the moles of methane

The molar - mass of methane ($CH_4$) is $M = 16.05\ g/mol$. The number of moles of methane, $n_{CH_4}=\frac{m}{M}=\frac{5.00\ g}{16.05\ g/mol}\approx0.312\ mol$

Step3: Determine the theoretical moles of water

From the balanced equation, the mole - ratio of $CH_4$ to $H_2O$ is $1:2$. So the theoretical moles of water, $n_{H_2O}^{theo}=2\times n_{CH_4}=2\times0.312\ mol = 0.624\ mol$

Step4: Calculate the theoretical mass of water

The molar - mass of water ($H_2O$) is $M = 18.02\ g/mol$. The theoretical mass of water, $m_{H_2O}^{theo}=n_{H_2O}^{theo}\times M=0.624\ mol\times18.02\ g/mol\approx11.24\ g$

Step5: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{m_{H_2O}^{actual}}{m_{H_2O}^{theo}}\times100%$. Substituting the actual mass of water ($m_{H_2O}^{actual}=6.10\ g$) and the theoretical mass of water ($m_{H_2O}^{theo}\approx11.24\ g$) into the formula, we get $\text{Percent Yield}=\frac{6.10\ g}{11.24\ g}\times100%\approx54.3%$

Answer:

$54.3$