Question

consider a reaction of combustion of methane that produces carbon dioxide and water. write a balanced equation (figure 7.3) and fill in missing numbers below. note to students: for molar mass values, use periodic table from the textbook. all answers must be reported with the correct number of significant figures. a) when 3 moles of methane react with 6 moles of oxygen gas moles of carbon dioxide and moles of water are produced. a) when 0.2 moles of methane react with 0.4 moles of oxygen gas moles of carbon dioxide and moles of water are produced.

Explanation:

Step1: Write the balanced combustion equation

The combustion of methane ($CH_4$) in oxygen ($O_2$) to produce carbon - dioxide ($CO_2$) and water ($H_2O$) is $CH_4 + 2O_2
ightarrow CO_2+2H_2O$. This means 1 mole of $CH_4$ reacts with 2 moles of $O_2$ to produce 1 mole of $CO_2$ and 2 moles of $H_2O$.

Step2: Determine the limiting reactant for the first case

Given 3 moles of $CH_4$ and 6 moles of $O_2$. The mole - ratio of $CH_4$ to $O_2$ from the balanced equation is 1:2. For 3 moles of $CH_4$, we would need $3\times2 = 6$ moles of $O_2$. Since the amount of $O_2$ is exactly what is required for the complete reaction of $CH_4$, neither reactant is in excess.
The moles of $CO_2$ produced: Using the mole - ratio of $CH_4$ to $CO_2$ (1:1), when 3 moles of $CH_4$ react, 3 moles of $CO_2$ are produced.
The moles of $H_2O$ produced: Using the mole - ratio of $CH_4$ to $H_2O$ (1:2), when 3 moles of $CH_4$ react, $3\times2=6$ moles of $H_2O$ are produced.

Step3: Determine the limiting reactant for the second case

Given 0.2 moles of $CH_4$ and 0.4 moles of $O_2$. The mole - ratio of $CH_4$ to $O_2$ from the balanced equation is 1:2. For 0.2 moles of $CH_4$, we need $0.2\times2 = 0.4$ moles of $O_2$. Since the amount of $O_2$ is exactly what is required for the complete reaction of $CH_4$, neither reactant is in excess.
The moles of $CO_2$ produced: Using the mole - ratio of $CH_4$ to $CO_2$ (1:1), when 0.2 moles of $CH_4$ react, 0.2 moles of $CO_2$ are produced.
The moles of $H_2O$ produced: Using the mole - ratio of $CH_4$ to $H_2O$ (1:2), when 0.2 moles of $CH_4$ react, $0.2\times2 = 0.4$ moles of $H_2O$ are produced.

Answer:

First case: 3, 6
Second case: 0.2, 0.4