Methanol, CH3OH, is a useful fuel that can be made as follows: CO (g) + 2H2 (g) → CH3OH (l) A reaction mixture used 12.0 g of H2 and 74.5 g of CO. (a) Determine the theoretical yield of CH3OH. (b) Calculate the amount of the excess reactant that remains unchanged at the end of the reaction.

A = Theoretical yield = 82.24 g

B = Amount of excess reactant left = 1.72 g

Explanation:

Given dа ta:

Mass of H₂ = 12 g

Mass of CO = 74.5 g

Theoretical yield of CH₃OH = ?

Amount of excess reactant left = ?

Solution:

First of all we will write the balance chemical equation:

CO + 2H₂ → CH₃OH

Number of moles of H₂ = mass / molar mass

Number of moles of H₂ = 12 g / 2 g/mol = 6 mol

Number of moles of CO = mass / molar mass

Number of moles of CO = 74.5 g / 29 g/mol = 2.57 mol

Now we compare the moles of CH₃OH with CO and H₂ from balance chemical equation:

CO : CH₃OH

1 : 1

2.57 : 2.57

H₂ : CH₃OH

2 : 1

6 : 1/2 * 6 = 3 mol

The number of moles of CH₃OH produce by CO are less so it will limiting reactant.

mass of CH₃OH = number of moles * molar mass

mass of CH₃OH = 2.57 mol * 32 g/mol

mass of CH₃OH = 82.24 g

Excess amount of H₂:

CO : H₂

1 : 2

2.57 : 2*2.57 = 5.14 mol

The moles of H₂ that react with CO are 5.14. While the total number of moles of H₂ available are 6 moles. So,

The number of moles of H₂ remain untreated = 6 mol - 5.14 mol = 0.86 mol

Mass of H₂ remain untreated = 0.86 mol * 2 g/mol

Mass of H₂ remain untreated = 1.72 g