a 5.0 mole sample of methane is combined with 8.0 moles of oxygen and undergoes combustion. what is the…

a 5.0 mole sample of methane is combined with 8.0 moles of oxygen and undergoes combustion. what is the theoretical yield of water?\n\nch₄ + 2o₂ → co₂ + 2h₂o\n\na 5.0 moles\nb 10. moles\nc 8.0 moles\nd 4.0 moles

a 5.0 mole sample of methane is combined with 8.0 moles of oxygen and undergoes combustion. what is the theoretical yield of water?\n\nch₄ + 2o₂ → co₂ + 2h₂o\n\na 5.0 moles\nb 10. moles\nc 8.0 moles\nd 4.0 moles

Answer

Explanation:

Step1: Determine the limiting reactant

The mole - ratio of $CH_4$ to $O_2$ from the balanced equation $CH_4 + 2O_2\rightarrow CO_2+2H_2O$ is $1:2$. For 5.0 moles of $CH_4$, the moles of $O_2$ required is $n_{O_2}=5.0\times2 = 10.0$ moles. But we have only 8.0 moles of $O_2$. For 8.0 moles of $O_2$, the moles of $CH_4$ required is $n_{CH_4}=\frac{8.0}{2}=4.0$ moles. So, $O_2$ is the limiting reactant.

Step2: Calculate the moles of water produced

The mole - ratio of $O_2$ to $H_2O$ is $2:2 = 1:1$ from the balanced equation. If 8.0 moles of $O_2$ react, the moles of $H_2O$ produced is also 8.0 moles according to the stoichiometry.

Answer:

C. 8.0 moles