Question

Methanol (CH3OH) can be synthesized by the following reaction: CO (g) + 2 H2 (g) → CH3OH (g)

What volume (in liters) of hydrogen gas, at a temperature of 355 K and a pressure of 98.4 kPa, is required to synthesize 35.7 g of methanol given an excess of CO gas?

1. A) 7.47 L
2. B) 15.6 L
3. C) 32.3 L
4. D) 66.8 L
5. E) 93.6 L

Answer 1

Methanol $(CH_3OH)$ can be synthesized as:

$CO+2H_2→CH_3OH$

Temperature = 355 K

Pressure = 738 mmHg = 738/760 × 101325 = 98391.9 N/m²

mass of CH₃OH = 35.7 g/mol

molar mass CH₃OH = 32.04 g/mol

number of moles of CH₃OH = 1.114 moles

For every two moles of H₂ gas that react, one mole of CH₃OH is produced.

This means that 2.228 moles of hydrogen gas are needed to react to produce 1.114 moles of CH₃OH. 

Using the Ideal gas Law,

$PV = nRT$

$V = \dfrac{nRT}P = \dfrac{2.228× 8.314 × 355}{98391.9}$

V = 0.066833m³ = 66.833L

$\therefore$ 66.833 L of H₂ is needed to react with Carbon monoxide to yield 35.7 grams of methanol.