Question

Calculate the mass of the solid produced formed when 1.5 L of 0.50 M NaOH is combined with 2.0 L 0.10 M MgCl2.

Answer 1

Solution:

Balanced chemical equation:

MgCl₂(aq) + 2NaOH(aq) → Mg(OH)₂(s) + 2NaCl(aq)

According to the equation above: n(MgCl₂) = n(NaOH)/2 = n(Mg(OH)₂)

1.5 L of a 0.50 M solution of NaOH contains:

n(NaOH) = concentration × volume = (0.50 mol/L) × (1.5 L) = 0.75 mol NaOH

2.0 L of a 0.10 M solution of MgCl₂ contains:

n(MgCl₂) = concentration × volume = (0.10 mol/L) × (2.0 L) = 0.20 mol MgCl₂

Determine which reagent is limiting:

if NaOH is limiting it would react with (0.75 mol / 2) = 0.375 mol of MgCl₂

The initial amount of MgCl₂ (0.20 mol) is less than this.

Therefore, MgCl₂ is limiting reagent.

Based on the amount of MgCl₂ present, the maximum amount of Mg(OH)₂ formed is:

n(Mg(OH)₂) = n(MgCl₂) = 0.20 mol

The molar mass of Mg(OH)₂ is 58.32 g/mol.

Therefore,

(0.20 mol Mg(OH)₂) × (58.32 g Mg(OH)₂ / 1 mol Mg(OH)₂) = 11.664 g Mg(OH)₂ = 11.7 g Mg(OH)₂

Answer: 11.7 grams of the solid produced.