Question

Methanol is produced industrially by catalytic hydrogenation of carbon monoxide according to the following equation: CO(g) + 2 H2(g) → CH3OH(l) If the yield of the reaction is 40%, what volume of CO (measured at STP) would be needed to produce 1.0 × 106 kg CH3OH?

Answer 1

Answer:

1.7 × 10⁹ L

Explanation:

Step 1: Write the balanced equation

CO(g) + 2 H₂(g) → CH₃OH(l)

Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

$1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol$

Step 3: Calculate the theoretical yield of CH₃OH

The real yield of CH₃OH is 3.1 × 10⁷ mol  and the percent yield is 40%. The theoretical yield is:

$3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)$

Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH

The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol

Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP

The volume of 1 mole of CO at STP is 22.4 L.

$7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L$