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
The theoretical yield of methanol is 85 g and 1.36 g of excess reactant is left over.
The equation of the reaction is; CO(g) + 2H2(g) → CH3OH(l)
Number of moles of H2 = 12.0 g/2.00 g/mol = 6 moles
Number of moles of CO = 74.5 g /28 g/mol = 2.66 moles
1 mole of CO reacts with 2 moles of H2
2.66 moles of CO reacts with 2.66 moles × 2 moles/1 mole
= 5.32 moles of H2
H2 is the reactant in excess because there is more than enough H2 to react with CO.
Hence;
1 mole of CO produces 1 mole of methanol
2.66 moles of CO produces 2.66 moles of methanol
Theoretical yield of methanol = 2.66 moles of methanol × 32 g/mol = 85 g
The excess reactant is H2. Amount of excess H2 is obtained from;
Amount of H2 present - amount of H2 reacted
6 .00 moles - 5.32 moles = 0.68 moles
Mass of unreacted H2 = 0.68 moles × 2.00 g/mol = 1.36 g of H2
Learn more: https://brainly.com/question/5803619
