Answer:
Hydrogen will left.
Mass of hydrogen left = 88 g
Mass of methane produced = 73.72 g
Explanation:
Given data:
Molar mass of methanol = 32.05 g/mol
Mass of carbon monoxide = 63.4 g
Molar mass of carbon monoxide = 28.01 g/mol
Mass of hydrogen gas = 97.2 g
Molar mass of hydrogen gas = 2 g/mol
Mass of methanol = ?
Excess reactant = ?
Amount of excess reactant left = ?
Solution:
Chemical equation:
CO(g) + 2H₂(g) → CH₃OH(g)
Number of moles of CO
Number of moles = mass/ molar mass
Number of moles = 63.4 g/ 28.01 g/mol
Number of moles = 2.3 mol
Number of moles of H₂:
Number of moles = mass/ molar mass
Number of moles = 97.2 g/ 2 g/mol
Number of moles = 48.6 mol
Now we compare the moles of methane with hydrogen and carbon monoxide gas through balanced chemical equation.
CO : CH₃OH
1 : 1
2.3 : 2.3
H₂ : CH₃OH
2 : 1
48.6 : 1/2×48.6 = 24.45 mol
Number of moles of methane produced by CO are less so it will limiting reactant.
Mass of methane = moles × molar mass
Mass of methane = 2.3 mol × 32.05 g/mol
Mass of methane = 73.72 g
The hydrogen gas is in excess amount. Some of its amount will remain at the end of reaction unreacted.
Amount of hydrogen remain unreacted:
CO : H₂
1 : 2
2.3 : 2×2.3 = 4.6 mole
so 4.6 moles of H₂ react with 2.3 moles of CO.
Total number of moles of hydrogen = 48.6 mole
Number of moles left = 48.6 - 4.6 = 44 moles
Mass of hydrogen left = moles × molar mass
Mass of hydrogen left = 44 mol × 2g/mol
Mass of hydrogen left = 88 g
