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To calculate the percent yield of methanol (CH₃OH), we follow these steps:

### Step 1: Determine the theoretical yield.
The balanced reaction is:
\[
\text{CO(g)} + 2 \text{H₂(g)} \rightarrow \text{CH₃OH(l)}
\]
This tells us that 2 moles of H₂ are needed to produce 1 mole of CH₃OH.

Given:
- 9 moles of H₂ are used.
- The ratio of H₂ to CH₃OH is 2:1, meaning for every 2 moles of H₂, 1 mole of CH₃OH is produced.

The theoretical yield of CH₃OH:
\[
\frac{9 \text{ moles of H₂}}{2} = 4.5 \text{ moles of CH₃OH (theoretical yield)}.
\]

### Step 2: Calculate the percent yield.
The actual yield of CH₃OH is given as 4 moles.

The formula for percent yield is:
\[
\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100
\]

Substitute the values:
\[
\text{Percent Yield} = \frac{4}{4.5} \times 100 = 88.9\%
\]

### Conclusion:
The percent yield of methanol is **88.9%**.

Thus, the correct answer is **D. 88.9%**.

Would you like further details or explanations? Here are five related questions to explore this topic further:

1. How is the theoretical yield calculated in a chemical reaction?
2. What factors can affect the actual yield of a chemical reaction?
3. Can the percent yield ever be greater than 100%?
4. How would the calculation change if the limiting reagent were CO instead of H₂?
5. Why is excess CO used in this industrial reaction?

**Tip:** Always ensure to balance the chemical equation before starting yield calculations to get accurate results.

Methanol can be produced industrially by using high pressure to carry out the following reaction: CO(g) + 2 H2(g) → CH3OH(l). Calculate the percent yield of methanol if 4 mol of CH3OH(l) are obtained from the reaction of 9 mol of H2(g) with excess CO(g).

Learn how to calculate the percent yield of methanol in a chemical reaction involving hydrogen gas (H2) and carbon monoxide (CO). This example covers stoichiometric ratios and the percent yield formula with a detailed step-by-step solution. Suitable for students in Grades 10-12.

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