What is the theoretical yield of NaCl would result from reacting 5.00 mil of Na2CO3 with excess HCl?

584.4 g.

Explanation:

Na₂CO₃ reacts with HCl according to the balanced reaction:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.

It is clear that 1.0 mole of Na₂CO₃ reacts with HCl to produce 2.0 moles of NaCl.

Using cross multiplication:

1.0 mole of Na₂CO₃ → 2.0 moles of NaCl.

5.0 moles of Na₂CO₃ →? moles of NaCl.

The no. of moles of NaCl produced from 2.0 moles of Na₂CO₃ = (2.0 mol) (5.0 mol) / (1.0 mol) = 10.0 mol.

∴ The theoretical yield of NaCl = no. of moles of NaCl x molar mass of NaCl = (10.0 mol) (58.44 g/mol) = 584.4 g.