Answer:
16.95 g of LiCl and 28.4 g of Na₂SO₄.
Step-by-step explanation:
- The reaction between NaCl and Li₂SO₄ is represented as:
2NaCl + Li₂SO₄ → 2LiCl + Na₂SO₄.
- it is clear that 2.0 moles of NaCl react with 1.0 mole of Li₂SO₄ to produce 2.0 mole of LiCl and 1.0 mole of Na₂SO₄.
- We need to calculate the no. of moles of NaCl (23.45 g) and Li₂SO₄ (56.23 g) using the relation:
n = mass/molar mass,
n of NaCl = mass/molar mass = (23.45 g)/(58.44 g/mol) = 0.40 mol.
n of Li₂SO₄ = mass/molar mass = (56.23 g)/(109.94 g/mol) = 0.51 mol.
- Since, every 1.0 mole of Li₂SO₄ reacts with 2.0 moles of NaCl, so 0.2 mole of Li₂SO₄ will react with 0.4 mol of NaCl.
∴ NaCl is the limiting reactant and Li₂SO₄ is in excess.
- So, the products will be LiCl (0.4 mol) and Na₂SO₄ (0.2 mol).
- Now, we can get the masses of each product using the relation:
mass = n x molar mass.
∴ mass of LiCl = n x molar mass = (0.4 mol)(42.394 g/mol) = 16.95 g.
∴ mass of Na₂SO₄ = n x molar mass = (0.2 mol)(142.04 g/mol) = 28.4 g.