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:

2NaCl + Li₂SO₄ → 2LiCl + Na₂SO₄.

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.

∴ NaCl is the limiting reactant and Li₂SO₄ is in excess.

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.

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.