What mass of methanol (CH3OH) is produced when 140.1 g of carbon monoxide reacts with 12.12 g of hydrogen?
CO(g) + 2H2(g) → CH2OH()
A. 160.2g
B. 96.12 g
C. 6.06 g
D. 192.2g

B. 96.12 g

Explanation:

To solve this question we must convert each reactant to moles. Using the balanced equation we can find limiting reactant. With moles of limiting reactant we can find the moles of methanol and its mass:

Moles CO -Molar mass: 28.01g/mol-

140.1g CO * (1mol / 28.01g) = 5.0 moles CO

Moles H2 -Molar mass: 2.01g/mol-

12.12g * (1mol / 2.01g) = 6.03 moles H2

For a complete reaction of 5.0 moles of CO are required:

5.0moles CO * (2mol H2 / 1mol CO) = 10.0 moles of H2 are required

As there are just 6.03 moles, H2 is limiting reactant.

The moles of methanol produced are:

6.03 moles H2 * (1mol CH3OH / 2mol H2) = 3.015 moles CH2OH

Mass CH3OH -Molar mass32.04g/mol-:

3.015 moles CH2OH * (32.04g/mol) =

96.6g CH3OH ≈ B. 96.12 g

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