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?

To answer these questions, we need to use the balanced chemical equation for the reaction:

CO(g) + 2H2(g) → CH3OH(g)

a) To determine the moles of each reactant needed to produce 3.60 x 10^2 g CH3OH, we can use molar mass and stoichiometry.

Molar mass of CH3OH = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol

Moles of CH3OH = mass/molar mass = 3.60 x 10^2 g / 32.04 g/mol = 11.23 mol

According to the balanced equation, 1 mole of CH3OH requires 1 mole of CO and 2 moles of H2. Therefore:

Moles of CO = 11.23 mol
Moles of H2 = 2 x 11.23 mol = 22.46 mol

b) To calculate the number of grams of each reactant needed to produce 4.00 mol CH3OH, we can use molar mass and stoichiometry.

Moles of CH3OH = 4.00 mol

Molar mass of CH3OH = 32.04 g/mol

Mass of CH3OH = moles x molar mass = 4.00 mol x 32.04 g/mol = 128.16 g

According to the balanced equation, 1 mole of CH3OH requires 1 mole of CO and 2 moles of H2. Therefore:

Moles of CO = 4.00 mol
Moles of H2 = 2 x 4.00 mol = 8.00 mol

To find the mass of CO and H2, we need to use their respective molar masses:

Molar mass of CO = 28.01 g/mol
Mass of CO = moles x molar mass = 4.00 mol x 28.01 g/mol = 112.04 g

Molar mass of H2 = 2.02 g/mol
Mass of H2 = moles x molar mass = 8.00 mol x 2.02 g/mol = 16.16 g

c) To determine the grams of hydrogen necessary to react with 2.85 mol CO, we use stoichiometry.

Moles of CO = 2.85 mol

According to the balanced equation, 1 mole of CO requires 2 moles of H2. Therefore:

Moles of H2 = 2 x 2.85 mol = 5.70 mol

To find the mass of H2, we need to use its molar mass:

Molar mass of H2 = 2.02 g/mol
Mass of H2 = moles x molar mass = 5.70 mol x 2.02 g/mol = 11.53 g

Therefore, 11.53 grams of hydrogen are necessary to react with 2.85 moles of CO.

To answer these questions, we can use the balanced chemical equation:

CO(g) + 2H2(g) ==> CH3OH(g)

a) To find the number of moles of each reactant needed to produce 3.60 x 10^2 g of CH3OH, we need to calculate the stoichiometric ratio.

The molar mass of CH3OH is 32.04 g/mol.

Using the equation, we can see that one mole of CH3OH is produced from one mole of CO and two moles of H2. Therefore, the stoichiometric ratio is:

1 mol CO : 2 mol H2 : 1 mol CH3OH

To calculate the number of moles of each reactant, we divide the given mass of CH3OH by its molar mass:

Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
Moles of CH3OH = 3.60 x 10^2 g / 32.04 g/mol

b) To calculate the number of grams of each reactant needed to produce 4.00 mol of CH3OH, we use the same stoichiometric ratio as before.

Using the equation, we can see that one mole of CH3OH is produced from one mole of CO and two moles of H2.

To calculate the grams of CO needed, we multiply the given moles of CH3OH by the molar mass ratio:

Grams of CO = Moles of CH3OH * (Molar mass of CO / Molar mass of CH3OH)
Grams of CO = 4.00 mol * (28.01 g/mol / 32.04 g/mol)

To calculate the grams of H2 needed, we multiply the given moles of CH3OH by the molar mass ratio:

Grams of H2 = Moles of CH3OH * (2 * Molar mass of H2 / Molar mass of CH3OH)
Grams of H2 = 4.00 mol * (2 * 2.02 g/mol / 32.04 g/mol)

c) To find the grams of hydrogen necessary to react with 2.85 moles of CO, we use the stoichiometric ratio.

Using the equation, we know that one mole of CO reacts with two moles of H2.

To calculate the grams of H2 needed, we multiply the given moles of CO by the molar mass ratio:

Grams of H2 = Moles of CO * (2 * Molar mass of H2 / Molar mass of CO)
Grams of H2 = 2.85 mol * (2 * 2.02 g/mol / 28.01 g/mol)

By following these calculations, you can find the number of moles or grams of each reactant required in a chemical reaction involving methanol production.

a) 3.6 x 10^2 g CH3OH = ?? moles.

3.5 x 10^2/molar mass CH3OH = moles.
You will need that many moles CO and twice that for moles H2.
b) To produce 4.00 mole CH3OH, you will need 4 moles CO and 8 moles H2. Convert those moles to grams. g = moles x molar mass.
c) Use the coefficients in the balanced equation to convert 2.85 mol CO to moles H2, then convert moles H2 to grams.