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 an

Question

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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 an

d 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.

Chemistry

1 answer:

algol [13] · Answer 1 · 2022-05-26T19:37:49+03:00

Answer:

A = Theoretical yield = 82.24 g

B = Amount of excess reactant left = 1.72 g

Explanation:

Given data:

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