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?
Methanol "(CH_3OH)" can be synthesized as:
"CO+2H_2\u2192CH_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\u00d7 8.314 \u00d7 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.
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