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We are given with the reaction that produces methanol from the reaction of hydrogen gas and carbon monoxide. This is expressed in the balanced equation: 2H2+CO=CH3OH. We need to identify the limiting reactant. Convert each mass to moles and divide each with their corresponding stoich. coeff. For H2, this is equal to 3 and for CO, this is equal to 2.66. Hence CO is the limiting reactant. From this, the amount of methanol produced is 85.14 grams.

Answer : The theoretical yield of the methanol is, 85.12 g

Solution : Given,

Mass of \(H_2\) = 12 g

Mass of \(CO\)= 74.5 g

Molar mass of \(H_2\) = 2 g/mole

Molar mass of \(CO\) = 28 g/mole

Molar mass of \(CH_3OH\) = 32 g/mole

First we have to calculate the moles of \(H_2\) and \(CO\).

Moles of \(H_2\) = \(\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{12g}{2g/mole}=6moles\)

Moles of \(CO\) = \(\frac{\text{ given mass of }CO}{\text{ molar mass of }CO}= \frac{74.5g}{28g/mole}=2.66moles\)

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

\(CO+2H_2\rightarrow CH_3OH\)

From the balanced reaction we conclude that

As, 1 mole of CO react with 2 moles of \(H_2\)

So, 2.66 moles of CO react with \(2\times 2.66=5.32moles\) of \(H_2\)

Excess moles of \(H_2\) = 6 - 5.32 = 0.68 moles

That means CO is a limiting reagent and \(H_2\) is an excess reagent.

Now we have to calculate the moles of methanol.

From the reaction we conclude that,

As, 1 mole of CO react to give 1 mole of methanol

So, 2.66 moles of CO react to give 2.66 moles of methanol

Now we have to calculate the mass of methanol.

\(\text{Mass of }CH_3OH=\text{Moles of }CH_3OH\times \text{Molar mass of }CH_3OH\)

\(\text{Mass of }CH_3OH=(2.66moles)\times (32g/mole)=85.12g\)

Therefore, the theoretical yield of the methanol is, 85.12 g

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