To determine the number of grams of methane ([tex]\(CH_4\)[/tex]) produced from the complete reaction of 3.75 moles of hydrogen gas ([tex]\(H_2\)[/tex]), we will follow these steps:
### Step 1: Understand the Reaction Stoichiometry
The balanced chemical equation is:
[tex]\[ CO + 3H_2 \rightarrow H_2O + CH_4 \][/tex]
According to the stoichiometry of the reaction:
- 3 moles of [tex]\( H_2 \)[/tex] produce 1 mole of [tex]\( CH_4 \)[/tex].
### Step 2: Calculate the Moles of Methane Produced
Given:
- 3.75 moles of [tex]\( H_2 \)[/tex]
We can use the mole ratio from the balanced equation (3 moles [tex]\( H_2 \)[/tex] : 1 mole [tex]\( CH_4 \)[/tex]):
[tex]\[
\text{Moles of } CH_4 = \frac{\text{Moles of } H_2}{3} = \frac{3.75 \, \text{moles} \, H_2}{3} = 1.25 \, \text{moles} \, CH_4
\][/tex]
### Step 3: Calculate the Grams of Methane Produced
To convert moles of [tex]\( CH_4 \)[/tex] to grams, we use the molar mass of [tex]\( CH_4 \)[/tex]:
- Molar mass of [tex]\( CH_4 \)[/tex] = 16.05 g/mol
Using the formula:
[tex]\[
\text{Grams of } CH_4 = \text{Moles of } CH_4 \times \text{Molar mass of } CH_4
\][/tex]
[tex]\[
\text{Grams of } CH_4 = 1.25 \, \text{moles} \, CH_4 \times 16.05 \, \text{g/mol} = 20.0625 \, \text{grams}
\][/tex]
### Final Answer
Therefore, the complete reaction of 3.75 moles of hydrogen gas ([tex]\(H_2\)[/tex]) produces 20.0625 grams of methane ([tex]\(CH_4\)[/tex]).
