Methanol is produced as the sole product of the combination of carbon monoxide gas and hydrogen gas. If 3.12X103 kg of methanol are actually produced, what is the precent yield if you start with 430.0 kg of hydrogen gas?
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To calculate the percent yield of methanol, we need to compare the actual yield (3.12 × 10^3 kg) to the theoretical yield, which is the amount of methanol that would be produced if the reaction went to completion.
First, let's calculate the amount of methanol that can be produced from 430.0 kg of hydrogen gas.
The balanced chemical equation for the reaction is:
CO + 2H2 → CH3OH
According to the equation, one mole of carbon monoxide (CO) reacts with two moles of hydrogen gas (H2) to produce one mole of methanol (CH3OH).
To convert the mass of hydrogen gas to moles, we need to know the molar mass of hydrogen (H2), which is approximately 2.016 g/mol.
So, the number of moles of hydrogen gas (H2) can be calculated as:
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 430.0 kg / (2.016 g/mol)
moles of H2 = 213.89 mol
According to the balanced equation, one mole of methanol (CH3OH) is produced for every two moles of hydrogen gas (H2) reacted.
Hence, the theoretical yield of methanol can be calculated as:
moles of methanol = moles of H2 / 2
moles of methanol = 213.89 mol / 2
moles of methanol = 106.945 mol
Lastly, we need to convert the moles of methanol to kilograms:
mass of methanol = moles of methanol × molar mass of methanol
molar mass of methanol = 32.04 g/mol
mass of methanol = 106.945 mol × (32.04 g/mol)
mass of methanol = 3428.15 g
mass of methanol = 3.43 × 10^3 kg
Now we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) × 100
percent yield = (3.12 × 10^3 kg / 3.43 × 10^3 kg) × 100
percent yield = 90.9%
Therefore, the percent yield of methanol in this reaction, starting with 430.0 kg of hydrogen gas, is approximately 90.9%.