Methanol (CH3OH) burns in air according to the equation 2CH3OH + 3O2 → 2CO2 + 4H2O If 201 g of methanol are used up in a combustion process, what is the mass of H2O produced?

The mass of water produced by 201 grams of methanol has been 451.685 grams.

According to the balanced chemical equation, the complete combustion of 2 moles of methanol produces 4 moles of water.

Moles can be calculated by the expression:

Moles =

The moles of methanol in 201 g has been:

Moles of methanol =

Moles of methanol = 6.273 mol.

From the equation:

1 mole methanol = 4 moles water

6.273 moles methanol = 6.273 × 4 moles water

6.273 moles methanol = 25.093 moles water.

The moles of water produced by 201 grams of methanol have been 25.093 mol.

The mass of water can be calculated as:

Mass = moles × molecular mass

Mass of water produced = 25.093 mol × 18 g/mol

Mass of water produced = 451.685 grams.

The mass of water produced by 201 grams of methanol has been 451.685 grams.

For more information about the mass-produced in a chemical reaction, refer to the link:

link

226.0 grams of H2O is produced

Explanation:

Step 1: Data given

Mass of methanol = 201 grams

Molar mass  methanol = 32.04 g/mol

Step 2: The balanced equation

2CH3OH + 3O2 → 2CO2 + 4H2O

Step 3: Calculate moles methanol

Moles methanol = mass methanol / molar mass

Moles methanol = 201.0 grams / 32.04 g/mol

Moles methanol = 6.27 moles

Step 4: Calculate moles H2O

For 2 moles methanol we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

For 6.27 moles methanol we'll have 2*6.27 = 12.54 moles H2O

Step 5: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 12.54 moles * 18.02 g/mol

Mass H2O = 226.0 grams

226.0 grams of H2O is produced