If 4.0 g of H₂ are made to react with excess CO, 31.7 g CH₃OH should be formed. If 28.0 g of CH₃OH are actually produced, what is the percent yield? CO(g) + 2 H2(g) CH3OH(l) *

87.5 % is the percent yield

Explanation:

We convert the mass of H₂ to moles: 4 g / 2g/mol = 2 moles of H₂

The reaction is: CO(g) + 2H₂(g) → CH₃OH(l)

So ratio is 2:1. Therefore, If I have 2 moles of hydrogen I would produce 1 mol of methanol.

So, as molar mass of methanol is 32g/mol; 32 g is the mass of produced methanol, at the 100 % yield reaction (theoretical yield)

Percent yield will be: (produced yield/ theoretical yield) . 100

(28 g / 32g) . 100 = 87.5 %

The percent yield of the reaction is 88.3 %

Explanation:

Step 1: Data given

Mass of H2 = 4.0 grams

Mass of CH3OH produced = 31.7 grams

Mass of CH3OH actually produced = 28.0 grams

Step 2: The balanced equation

CO(g) + 2 H2(g) → CH3OH(l)

Step 3: Calculate the percent yield

Percent yield = (actual mass CH3OH / theoretical mass CH3OH) * 100 %

Percent yield = (28.0 grams / 31.7 grams) * 100%

Percent yield = 88.3 %

The percent yield of the reaction is 88.3 %

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