Hydrogen gas reacts with carbon monoxide to produce methane as shown in the balanced equation below.2 H2(g) + CO(g) ⟶⟶ CH3OH(l)In an experiment, 8.00 g of H2 is treated with 74.5 g of CO. If CH3OH is produced in a 94.0% yield, what mass of CH3OH is produced?Molar Masses:H2: 2.016 g/molCO: 28.01 g/molCH3OH: 32.042 g/mol 63.6 g 80.1 g 119.1 g 59.8 g 239.0 g
Question
Hydrogen gas reacts with carbon monoxide to produce methane as shown in the balanced equation below.2 H2(g) + CO(g) ⟶⟶ CH3OH(l)In an experiment, 8.00 g of H2 is treated with 74.5 g of CO. If CH3OH is produced in a 94.0% yield, what mass of CH3OH is produced?Molar Masses:H2: 2.016 g/molCO: 28.01 g/molCH3OH: 32.042 g/mol 63.6 g 80.1 g 119.1 g 59.8 g 239.0 g
Solution
First, we need to determine the number of moles for both H2 and CO.
For H2: 8.00 g / 2.016 g/mol = 3.97 mol
For CO: 74.5 g / 28.01 g/mol = 2.66 mol
From the balanced equation, we can see that the reaction is in a 2:1 ratio (2 moles of H2 for every mole of CO). Therefore, CO is the limiting reactant because we have less than the required 2:1 ratio.
Next, we calculate the theoretical yield of CH3OH. Since the reaction is in a 1:1 ratio with CO, the number of moles of CH3OH that should be produced is the same as the number of moles of CO, which is 2.66 mol.
To find the mass, we multiply by the molar mass of CH3OH: 2.66 mol * 32.042 g/mol = 85.2 g
However, the problem states that the yield is only 94.0%, so we need to adjust for this: 85.2 g * 0.94 = 80.1 g
Therefore, the mass of CH3OH produced is 80.1 g.
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