To find the mass of methanol produced, we compare the number of moles of each reactant to the stoichiometry of the reaction. H2 is the limiting reactant, so the mass of methanol produced is equal to the molar mass of methanol multiplied by the number of moles of H2. The mass of the excess reactant is 0 g.
Step-by-step explanation:
To find the mass of methanol that can be produced, we first need to determine which reactant is limiting. In order to do this, we compare the number of moles of each reactant to the stoichiometry of the reaction. We are given the masses of CO and H2, so we need to convert them to moles using their molar masses. CO has a molar mass of 28.01 g/mol, so 373 g of CO is equal to 373/28.01 = 13.31 mol. H2 has a molar mass of 2.02 g/mol, so 70.2 g of H2 is equal to 70.2/2.02 = 34.65 mol. According to the balanced equation, the ratio of CO to CH3OH is 1:1, and the ratio of H2 to CH3OH is 2:1. Therefore, we have an excess of H2, which means it is the limiting reactant.
To calculate the mass of methanol produced, we need to determine the number of moles of methanol formed. Since the limiting reactant is H2, the number of moles of methanol formed is equal to the number of moles of H2. Therefore, the mass of methanol produced is equal to the molar mass of methanol multiplied by the number of moles of methanol. The molar mass of methanol is 32.04 g/mol (1 carbon atom + 4 hydrogen atoms + 1 oxygen atom = 12.01 + 4(1.01) + 16.00 = 32.04 g/mol). So the mass of methanol produced is 32.04 * 34.65 = 1111.47 g.
To find the mass of the excess reactant, we need to calculate the mass of H2 that remains after the reaction. The number of moles of H2 consumed in the reaction is equal to the number of moles of methanol formed, which is 34.65 mol. Therefore, the number of moles of H2 remaining is equal to the initial number of moles of H2 minus the moles consumed in the reaction, which is 34.65 mol – 34.65 mol = 0 mol. Since the molar mass of H2 is 2.02 g/mol, the mass of the excess H2 is 0 mol * 2.02 g/mol = 0 g.