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) + 2H2(g) ==> CH3OH(g)
a) How many moles of each reactant are needed to produce 3.60 x 10^2 g CH3OH?
b)Calculate the number of grams of each reactant needed to produce 4.00 mol CH3OH.
c)How many grams of hydrogen are necessary to react with 2.85 mol CO?
a) To find the number of moles of each reactant needed to produce 3.60 x 10^2 g CH3OH, we need to use the molar mass of CH3OH.
The molar mass of CH3OH = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) + 1.01 g/mol (H) = 32.04 g/mol
So, to find the moles of CH3OH, we divide the mass of CH3OH by its molar mass:
Moles of CH3OH = (3.60 x 10^2 g CH3OH) / (32.04 g/mol) = 11.23 mol CH3OH
According to the balanced equation, the ratio of CO to CH3OH is 1:1. Therefore, the moles of CO required is also 11.23 mol.
The ratio of H2 to CH3OH is 2:1. Therefore, the moles of H2 required are twice the moles of CH3OH.
Moles of H2 = 2 x 11.23 mol = 22.46 mol
So, 11.23 mol of CO and 22.46 mol of H2 are needed to produce 3.60 x 10^2 g of CH3OH.
b) To calculate the number of grams of each reactant needed to produce 4.00 mol of CH3OH, we again use the molar mass of CH3OH.
Molar mass of CH3OH = 32.04 g/mol
The moles of CH3OH given is 4.00 mol.
Grams of CH3OH = (4.00 mol) x (32.04 g/mol) = 128.16 g CH3OH
According to the balanced equation, since the ratio of CO to CH3OH is 1:1, the number of grams of CO required is also 128.16 g.
The ratio of H2 to CH3OH is 2:1. Therefore, the grams of H2 required are twice the grams of CH3OH.
Grams of H2 = 2 x 128.16 g = 256.32 g
So, 128.16 g of CO and 256.32 g of H2 are needed to produce 4.00 mol of CH3OH.
c) To find the grams of hydrogen necessary to react with 2.85 mol of CO, we need to use the balanced equation to determine the mole ratio between CO and H2.
According to the balanced equation, the mole ratio is 1:2. This means that for every 1 mole of CO, we need 2 moles of H2.
Therefore, the mole ratio of CO to H2 is 1:2.
Given that there are 2.85 mol of CO, we multiply this by 2 to find the moles of H2:
Moles of H2 = 2.85 mol CO x (2 mol H2 / 1 mol CO) = 5.70 mol H2
To find the grams of H2, we multiply the moles of H2 by the molar mass of H2:
Grams of H2 = 5.70 mol H2 x (2.02 g/mol) = 11.54 g H2
So, 11.54 grams of hydrogen are necessary to react with 2.85 mol of CO.
a) To determine the number of moles of each reactant needed to produce 3.60 x 10^2 g of CH3OH, we need to use the molar mass of methanol.
The molar mass of CH3OH is:
(1 x 12.01 g/mol) + (4 x 1.01 g/mol) + (16.00 g/mol) = 32.04 g/mol
To find the number of moles of CH3OH, we divide the given mass of CH3OH by its molar mass:
Number of moles = Mass / Molar mass
Number of moles = 3.60 x 10^2 g / 32.04 g/mol
Number of moles = 11.233 mol
According to the balanced chemical equation, we have a 1:2:1 ratio between CO, H2, and CH3OH. Therefore, the number of moles of CO and H2 needed would be the same as the number of moles of CH3OH.
So, the number of moles of CO and H2 needed to produce 3.60 x 10^2 g of CH3OH is 11.233 mol each.
b) To calculate the number of grams of each reactant needed to produce 4.00 mol of CH3OH, we again use the molar mass of CH3OH to convert from moles to grams.
Number of grams = Number of moles * Molar mass
For CO:
Number of grams of CO = 4.00 mol * (1 * 12.01 g/mol) = 48.04 g
For H2:
Number of grams of H2 = 4.00 mol * (2 * 1.01 g/mol) = 8.08 g
c) To find the grams of hydrogen required to react with 2.85 mol of CO, we need to use the stoichiometry of the balanced chemical equation.
According to the balanced chemical equation, the ratio between CO and H2 is 1:2. This means that for every 1 mole of CO, we need 2 moles of H2.
So, to find the grams of H2 needed, we multiply the number of moles of CO by the molar mass of CO and then multiply by the ratio of H2 to CO in terms of moles.
Number of grams of H2 = 2.85 mol CO * (1 * 2 * 1.01 g/mol) = 5.772 g
a) To determine the number of moles of each reactant needed to produce 3.60 x 10^2 g CH3OH, we need to use the molar mass of CH3OH and the stoichiometry of the reaction.
The molar mass of CH3OH is:
(1 x atomic mass of C) + (4 x atomic mass of H) + (1 x atomic mass of O)
= (1 x 12.01 g/mol) + (4 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 32.04 g/mol
Now, using the stoichiometry of the reaction, we can determine the number of moles of CO and H2 required.
From the balanced equation, we see that the ratio of CO to CH3OH is 1:1, and the ratio of H2 to CH3OH is 2:1. Therefore, we only need to consider the ratio for H2.
Moles of H2 = (moles of CH3OH) x (2 moles of H2 / 1 mole of CH3OH)
= (3.60 x 10^2 g CH3OH) / (32.04 g/mol) x (2 mol H2 / 1 mol CH3OH)
b) To calculate the number of grams of each reactant needed to produce 4.00 mol CH3OH, we can use the same approach.
Moles of CO = (moles of CH3OH) x (1 mole of CO / 1 mole of CH3OH)
= 4.00 mol CH3OH x (1 mol CO / 1 mol CH3OH)
Moles of H2 = (moles of CH3OH) x (2 moles of H2 / 1 mole of CH3OH)
= 4.00 mol CH3OH x (2 mol H2 / 1 mol CH3OH)
c) To calculate the grams of hydrogen necessary to react with 2.85 mol CO, we again use the stoichiometry of the reaction.
Grams of H2 = (moles of CO) x (2 moles of H2 / 1 mole of CO) x (molar mass of H2)
Note: The molar mass of H2 is simply the atomic mass of hydrogen, which is 1.01 g/mol.
Now, you can substitute the given values into the equations to calculate the answers.
lol its 3.45
Convert 360 g CH3OH to moles. moles = g/molar mass = about 360/32 about 11 moles but you need to do it more accurately than my estimates.
You will need 11 moles CH3OH x (1 mole CO/1 mole CH3OH) = 11 x 1/1 = about 11 moles CO
11 moles CH3OH x (2 moles H2/1 mole CH3OH) = 11 x 2/1 = about 22 moles H2.
b. Convert moles to grams. g = moles x molar mass
c. 2.85 mol CO x (2 moles H2/1 mol CO = 2.85 x (2/1) = about 5.7 moles H2. Convert to grams.