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?

221.6 g of H₂O are produced by this reaction

Explanation:

The combustion is this:

2CH₃OH + 3O₂ → 2CO₂ + 4H₂O

As we use 201 g of methanol, we assume the oxygen in excess.

2 moles of methanol can produce 4 moles of water.

We convert the grams of methanol to moles → 201 g. 1mol/32 g = 6.28 moles

Let's make a rule of three:

2 moles of methanol can produce 4 moles of water.

6.28 moles of methanol must produce (6.28.4) / 2 = 12.5 moles of H₂O

We convert the moles to mass → 12.5 mol. 18 g / 1mol = 226.1 g

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