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Certain race cars use methanol (CH 3 OH; also called wood alcohol) as a fuel. Me

ID: 890083 • Letter: C

Question

Certain race cars use methanol (CH3OH; also called wood alcohol) as a fuel. Methanol has a molecular mass of 32.0 g/mol and a density of 0.79 g/mL. The combustion of methanol occurs according to the following equation:

2CH3OH + 3O2 2CO2 + 4H2O

In a particular reaction 2.00 L of methanol are reacted with 80.0 kg of oxygen.

a.What is the limiting reactant?

b.What reactant and how many grams of it are left over?

c.How many grams of carbon dioxide is produced?

Honestly I am at a total loss on what to do.

Explanation / Answer

Certain race cars use methanol (CH3OH; also called wood alcohol) as a fuel. Methanol has a molecular mass of 32.0 g/mol and a density of 0.79 g/mL. The combustion of methanol occurs according to the following equation:

2CH3OH + 3O2 2CO2 + 4H2O

In a particular reaction 2.00 L of methanol are reacted with 80.0 kg of oxygen.

Solution :-

Lets calculate the mass of the methanol using the volume and density

2.00 L * 1000 ml / 1 L = 2000 ml

Mass = volume * density

         = 2000 ml * 0.79 g /ml

Now lets calculate the moles of the methanol and O2

Moles = mass / molar mass

Moles of methanol   = 1580 g methanol /32.0 g per mol

                                 = 49.375 mol methanol

Moles of O2 = (80.0 kg * 1000 g / 1 kg)*(1 mol O2/ 32 g ) = 2500 mol O2

Now lets calculate the moles of O2 needed to react with 49.375 mol methanol

Mole ratio is 2 mole methanol = 3 mol O2

49.375 mol methanol * 3 mol O 2 / 2 mol methanol = 74.0625 mol O2

Moles of O2 are more than needed for the reaction

Therefore methanol is the limiting reactant

Moles of O2 remain after reaction = 2500 mol – 74.0625 mol = 2425.9375 mol O2

Mass of O2 remain after reaction is

Mass of O2 remain = (2425.9375 mol * 32 g) =77630 g O2

Now lets calculate the mass of CO2 produced

Mole ratio of the methanol to CO2 is 2:2

49.375 mol methanol * 2 mol CO2/ 2 mol methanol = 49.375 mol CO2

Mass of CO2 = moles * molar mass

                     = 49.375 mol *44.01 g per mol

                     = 2173 g CO2

Therefore

a.What is the limiting reactant?

Methanol (CH3OH) is the limiting reactant

b.What reactant and how many grams of it are left over?

O2 reactant is left over and its mass is 77630 g

c.How many grams of carbon dioxide is produced?

                                Mass of CO2 produced = 2173 g