2= Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol
3= Mass of hydrogen = 0.45 g
3= Mass of oxygen = 3.616 g
4 = Theoretical yield of P₂O₅ = 14.2 g.
Explanation:
Q2= Given data:
Moles of PbS = 0.356 mol
Moles of PbO = ?
Moles of SO₂ = ?
Solution:
Chemical equation:
2PbS + 3O₂ → 2PbO + 2SO₂
Now we will compare the moles of PbS with PbO and SO₂ from balanced chemical equation.
PbS : PbO
2 : 2
0.356 : 0.356
Moles of PbO produced from 0.356 moles of PbS = 0.356 mol
PbS : SO₂
2 : 2
0.356 : 0.356
Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol
Q= 3
Given data:
Mass of water = 4.05 g
Mass of hydrogen = ?
Mass of oxygen = ?
Solution:
Chemical equation:
2H₂O → 2H₂ + O₂
Number of moles of water:
Number of moles of water = mass/ molar mass
Number of moles of water = 4.05 g/ 18 g/mol
Number of moles of water = 0.225 mol
Now we compare the moles of water with hydrogen and oxygen.
H₂O : H₂
2 : 2
0.225 : 0.225
Mass of hydrogen:
Mass of hydrogen = moles × molar mass
Mass of hydrogen = 0.225 mol × 2g/mol
Mass of hydrogen = 0.45 g
H₂O : O₂
2 : 1
0.225 : 1/2×0.225 = 0.113 mol
Mass of oxygen:
Mass of oxygen = moles × molar mass
Mass of oxygen = 0.113 mol × 32g/mol
Mass of oxygen = 3.616 g
Q 4
Given data:
Mass of phosphorus = 3.07 g
Mass of oxygen = 6.09 g
Theoretical yield of P₂O₅ = ?
Chemical equation:
4P + 5O₂ → 2P₂O₅
Number of moles of phosphorus = mass/ molar mass
Number of moles of phosphorus = 3.07 g/ 31 g/mol
Number of moles of phosphorus = 0.1 mol
Number of moles of oxygen = mass/ molar mass
Number of moles of oxygen = 6.09 g/ 32 g/mol
Number of moles of oxygen = 0.2 mol
Now we will compare the moles of P₂O₅ with oxygen and phosphorus.
O₂ : P₂O₅
5 : 2
0.2 : 2/5 ×0.2 = 0.08 mol
P : P₂O₅
4 : 2
0.1 : 2/4×0.1 = 0.05 mol
The number of moles of P₂O₅ produced from phosphorus are less that's why phosphorus will be limiting reactant.
Mass of P₂O₅ = moles × molar mass
Mass of P₂O₅ = 0.05 mol × 283.88 g/mol
Mass of P₂O₅ = 14.2 g
Theoretical yield of P₂O₅ = 14.2 g.
