If you have 23.45 g of NaCl and react it with 56.23 g of Li2 (SO4) How many grams of each product can be made?

16.95 g of LiCl and 28.4 g of Na₂SO₄.

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.