Question

For the following reaction, if 19.7 g of H2, is reacted with excess CO in the laboratory, and 144.5 g of CH3OH is produced, what is the percentage yield of CH3OH? Round your answer to the nearest whole percent.CO + 2 H2 → CH3OHSelect one:a.100%b.44 %c.82 %d.93 %

Answer 1

Step 1

The reaction involved:

CO + 2 H2 → CH3OH (completed and balanced)

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Step 2

Data provided:

19.7 g H2 (the limiting reactant)

Excess reactant = CO

144.5 g CH3OH = actual yield

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Data needed:

The molar masses of:

H2) 2.00 g/mol

CH3OH) 32.0 g/mol

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Step 3

The theoretical yield:

By stoichiometry,

CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)

2 x 2.00 g H2 --------- 32.0 g CH3OH

19.7 g H2 --------- X

X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2

X = 157.6 g CH3OH (The theoretical yield)

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Step 4

The % yield is defined as follows:

$\begin{gathered} \text{ \%yield = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}}x100\text{ } \\ \text{ \%yield = }\frac{144.5\text{ g}}{157.6\text{ g}}x100\text{ = 91.7 \% = 92 \% approx.} \end{gathered}$

Answer: d. 93% (it is the nearest value in comparison to my result)