Methanol (CH3OH) is used in the production of many chemicals. Methanol is made by reacting carbon monoxide and hydrogen at high temperature and pressure.
CO(g) + 2 H2(g) CH3OH(g)
(a) How many moles of each reactant are needed to produce 4.83 102 g CH3OH?
CO
mol
H2
mol
(b) Calculate the number of grams of each reactant needed to produce 3.37 mol CH3OH.
CO
H2
(c) How many grams of hydrogen are necessary to react with 2.45 mol CO?
H2
See your post above.
To solve this problem, we will use the stoichiometry of the reaction. The stoichiometric coefficients in the balanced equation represent the mole ratio between the reactants and products.
(a) To determine the number of moles of each reactant needed to produce 4.83 * 102 g CH3OH, we'll follow the steps below:
1. Calculate the molar mass of CH3OH:
Molar mass of CH3OH = (1*12.01) + (4*1.01) + 16.00 = 32.04 g/mol
2. Convert the given mass of CH3OH into moles:
moles of CH3OH = mass of CH3OH / molar mass of CH3OH
moles of CH3OH = 4.83 * 102 g / 32.04 g/mol
moles of CH3OH = 15.05 mol
3. Using the stoichiometric coefficients, determine the moles of CO and H2 based on the mole ratio in the balanced equation:
From the balanced equation: CO(g) + 2 H2(g) -> CH3OH(g)
1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH.
moles of CO = 15.05 mol
moles of H2 = 2 * 15.05 mol = 30.10 mol
Therefore, for the production of 4.83 * 102 g CH3OH, you would need 15.05 mol of CO and 30.10 mol of H2.
(b) To calculate the number of grams of each reactant needed to produce 3.37 mol CH3OH, follow these steps:
1. Given the moles of CH3OH, calculate the moles of CO and H2 using the stoichiometric coefficients:
From the balanced equation: CO(g) + 2 H2(g) -> CH3OH(g)
1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH.
moles of CH3OH = 3.37 mol
moles of CO = 3.37 mol
moles of H2 = 2 * 3.37 mol = 6.74 mol
2. Convert the moles of each reactant into grams using their respective molar masses:
mass of CO = moles of CO * molar mass of CO
mass of H2 = moles of H2 * molar mass of H2
Given that the molar mass of CO is 28.01 g/mol and the molar mass of H2 is 2.02 g/mol:
mass of CO = 3.37 mol * 28.01 g/mol
mass of H2 = 6.74 mol * 2.02 g/mol
Therefore, to produce 3.37 mol CH3OH, you would need 94.337 g of CO and 13.624 g of H2.
(c) To determine the grams of hydrogen necessary to react with 2.45 mol CO, follow these steps:
1. Given the moles of CO, we need to use the stoichiometric coefficients to determine the moles of H2 required.
From the balanced equation: CO(g) + 2 H2(g) -> CH3OH(g)
1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH.
moles of CO = 2.45 mol
2. Use the stoichiometry to find the moles of H2:
moles of H2 = 2 * moles of CO
moles of H2 = 2 * 2.45 mol = 4.90 mol
3. Convert the moles of H2 into grams using the molar mass of H2:
mass of H2 = moles of H2 * molar mass of H2
Given that the molar mass of H2 is 2.02 g/mol:
mass of H2 = 4.90 mol * 2.02 g/mol
Therefore, to react with 2.45 mol CO, you would need 9.898 g of H2.
To find the number of moles of each reactant needed to produce a certain amount of CH3OH, you can use the coefficients in the balanced chemical equation.
(a) To find the number of moles of CO and H2 needed to produce 4.83 * 10^2 g of CH3OH, you need to use the molar mass of CH3OH and the stoichiometric ratio between CH3OH and the reactants.
The molar mass of CH3OH can be calculated by adding the atomic masses of each element present in it:
C = 12.01 g/mol
H = 1.01 g/mol (3 hydrogen atoms)
O = 16.00 g/mol
Adding these together, the molar mass of CH3OH is 32.04 g/mol.
Now, you can set up the conversion factors to find the moles of CO and H2:
1 mol CH3OH = 32.04 g CH3OH
Using the stoichiometric ratio from the balanced equation:
1 mol CH3OH = 1 mol CO
1 mol CH3OH = 2 mol H2
To calculate the moles of CO and H2, divide the given mass of CH3OH by the molar mass:
Moles of CH3OH = (4.83 * 10^2 g) / (32.04 g/mol) = 15.06 mol CH3OH
Using the stoichiometric ratios, we can determine the moles of CO and H2 needed:
Moles of CO = 15.06 mol CH3OH * 1 mol CO / 1 mol CH3OH
Moles of CO = 15.06 mol CO
Moles of H2 = 15.06 mol CH3OH * 2 mol H2 / 1 mol CH3OH
Moles of H2 = 30.12 mol H2
Therefore, you would need 15.06 moles of CO and 30.12 moles of H2 to produce 4.83 * 10^2 g of CH3OH.
(b) To find the number of grams of each reactant needed to produce 3.37 mol of CH3OH, you can again use the stoichiometric ratios from the balanced equation.
Using the conversion factors:
1 mol CH3OH = 32.04 g CH3OH
1 mol CH3OH = 1 mol CO
1 mol CH3OH = 2 mol H2
Multiply the moles of CH3OH by the molar mass to get the grams of CH3OH:
Grams of CH3OH = 3.37 mol CH3OH * 32.04 g/mol = 107.93 g CH3OH
Now, using the stoichiometric ratios:
Grams of CO = 3.37 mol CH3OH * 1 mol CO / 1 mol CH3OH
Grams of CO = 3.37 mol CO * molar mass of CO
Grams of H2 = 3.37 mol CH3OH * 2 mol H2 / 1 mol CH3OH
Grams of H2 = 3.37 mol H2 * molar mass of H2
To calculate the grams of CO and H2, you need to know their molar masses. The molar mass of CO is 28.01 g/mol and the molar mass of H2 is 2.02 g/mol.
Therefore, the grams of CO needed would be 3.37 mol CO * 28.01 g/mol and the grams of H2 would be 3.37 mol H2 * 2.02 g/mol.
(c) To determine the grams of hydrogen necessary to react with 2.45 mol of CO, you can again use the stoichiometric ratio from the balanced equation.
Using the conversion factors:
1 mol CO = 28.01 g CO
1 mol CO = 1 mol CH3OH
1 mol CH3OH = 2 mol H2
Multiply the moles of CO by the molar mass to get the grams of CO:
Grams of CO = 2.45 mol CO * 28.01 g/mol
Now, using the stoichiometric ratio:
Grams of H2 = 2.45 mol CO * 2 mol H2 / 1 mol CO
Grams of H2 = 2.45 mol H2 * molar mass of H2
To calculate the grams of H2, you need to know its molar mass, which is 2.02 g/mol.
Therefore, the grams of H2 needed would be 2.45 mol H2 * 2.02 g/mol.