Answer:
Explanation:
Q 1;
Given data:
Mass of Na₂CO₃ produced = ?
Mass of sodium oxide = 50.0 g
Solution:
Chemical equation:
2Na₂O + CO → Na₂CO₃ + 2Na
Number of moles of sodium oxide:
Number of moles = mass/ molar mass
Number of moles = 50 g/ 62 g/mol
Number of moles = 0.806 mol
Now we will compare the moles of sodium oxide with sodium carbonate.
Na₂O : Na₂CO₃
2 : 1
0.806 : 1/2×0.81 = 0.403
Mass of sodium carbonate:
Mass = number of moles × molar mass
Mass = 0.403 mol × 106 g/mol
Mass = 42.7 g
Q 2;
Given data:
Moles of Al₂(SO₄)₃ produced = ?
Mass of Al = 35 g
Solution:
Chemical equation:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
Number of moles of Aluminium:
Number of moles = mass/ molar mass
Number of moles = 35 g/ 27 g/mol
Number of moles = 1.3 mol
Now we will compare the moles of sodium oxide with sodium carbonate.
Al : Al₂(SO₄)₃
2 : 1
1.3 : 1/2×1.3 = 0.65
So from 35 gram of Al 0.65 moles of Al₂(SO₄)₃ are produced in the presence of excess sulfuric acid.
