Final answer:
To produce 100 g of methanol (CH3OH) from CO, you would need 87.37 g of CO. The molar ratio between CO and CH3OH is 1:1. First, calculate the molar mass of CH3OH. Then, use the molar mass of CH3OH to calculate the moles of CH3OH in 100 g of methanol. Finally, use the molar ratio to find the mass of CO needed.
Step-by-step explanation:
To determine the mass of CO needed to produce 100 g of methanol (CH3OH), we need to use the balanced equation: CO(g) + H2(g) → CH3OH(g). From the equation, we can see that the molar ratio between CO and CH3OH is 1:1. First, we need to find the molar mass of CH3OH: 1 carbon atom (C) has a molar mass of 12.01 g/mol, 4 hydrogen atoms (H) have a total molar mass of 4.02 g/mol (1.01 g/mol per H atom), and 1 oxygen atom (O) has a molar mass of 16.00 g/mol. Therefore, the molar mass of CH3OH is 12.01 + 4.02 + 16.00 = 32.03 g/mol.
Next, we can use the molar mass of CH3OH to calculate the moles of CH3OH in 100 g of methanol. Moles of CH3OH = Mass of CH3OH (g) / Molar mass of CH3OH (g/mol) = 100 g / 32.03 g/mol = 3.12 mol of CH3OH.
Since the molar ratio between CO and CH3OH is 1:1, we need the same amount of moles of CO to produce 3.12 mol of CH3OH. Therefore, the mass of CO needed is equal to the molar mass of CO (28.01 g/mol) multiplied by the number of moles of CH3OH (3.12 mol): Mass of CO = Molar mass of CO (g/mol) * Moles of CH3OH = 28.01 g/mol * 3.12 mol = 87.37 g.