Let's go through the problem step by step to find out how many grams of methanol are produced when 2.9 grams of carbon monoxide reacts with 0.50 grams of hydrogen gas.
### Step 1: Calculate the moles of each reactant
First, we need the molar masses of the reactants and products:
- Molar mass of CO (Carbon Monoxide) = 28 g/mol
- Molar mass of H₂ (Hydrogen Gas) = 2 g/mol
- Molar mass of CH₃OH (Methanol) = 32 g/mol
#### Moles of CO:
[tex]\[
\text{Moles of CO} = \frac{\text{Mass of CO}}{\text{Molar Mass of CO}} = \frac{2.9 \, \text{g}}{28 \, \text{g/mol}} \approx 0.10357 \, \text{moles}
\][/tex]
#### Moles of H₂:
[tex]\[
\text{Moles of H₂} = \frac{\text{Mass of H₂}}{\text{Molar Mass of H₂}} = \frac{0.50 \, \text{g}}{2 \, \text{g/mol}} = 0.25 \, \text{moles}
\][/tex]
### Step 2: Determine the limiting reactant
The given reaction is:
[tex]\[
CO + 2 H_2 \rightarrow CH_3OH
\][/tex]
This tells us that 1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH.
From our calculations:
- Moles of CO = 0.10357
- Moles of H₂ = 0.25
Now we check if we have enough H₂ to react with all the CO:
[tex]\[
\frac{0.25 \, \text{moles of H₂}}{2} = 0.125 \, \text{moles CO needed}
\][/tex]
Since 0.10357 moles of CO is less than 0.125 moles (the moles of CO needed), CO is the limiting reactant.
### Step 3: Calculate the moles of methanol produced
Given that CO is the limiting reactant and reacts in a 1:1 molar ratio with CH₃OH, the moles of CH₃OH produced will be equal to the moles of CO:
[tex]\[
\text{Moles of CH₃OH} = \text{Moles of CO} = 0.10357
\][/tex]
### Step 4: Calculate the grams of methanol produced
Finally, we convert moles of methanol to grams:
[tex]\[
\text{Mass of CH₃OH} = \text{Moles of CH₃OH} \times \text{Molar Mass of CH₃OH} = 0.10357 \, \text{moles} \times 32 \, \text{g/mol} \approx 3.3143 \, \text{grams}
\][/tex]
Thus, the mass of methanol produced is approximately 3.31 grams. None of the given answer choices exactly match this value, but the closest option is 3.2 grams.
