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
