Question

8g of aqueous Sodium Hydroxide reacts with 4g of aqueous Aluminum Chloride to produce aqueous Sodium Chloride and solid Aluminum Hydroxide. Balance the equation and give the limiting and excess reactants.

Answer 1

Answer:

The limiting reactant is AlCl₃ and the excess reactant is NaOH.

Explanation:

• The balanced equation for the mentioned reaction is:

3NaOH(aq) + AlCl₃(aq) → 3NaCl(aq) + Al(OH)₃(s)↓,

It is clear that 3.0 moles of NaOH(aq) react with 1.0 mole of AlCl₃(aq) to produce 3.0 moles of NaCl(aq) and 1.0 mole of Al(OH)₃(s).

• Firstly, we need to calculate the no. of moles of (8.0 g) of NaOH and (4.0 g) of AlCl₃:

no. of moles of NaOH = mass/molar mass = (8.0 g)/(40.0 g/mol) = 0.2 mol.

no. of moles of AlCl₃ = mass/molar mass = (4.0 g)/(133.34 g/mol) = 0.03 mol.

• From stichiometry; NaOH reacts with AlCl₃ with (3: 1) molar ratio.

∴ 0.09 mol of NaOH (the remaining 1.1 mol is in excess) reacts completely with 0.03 mol of AlCl₃.

• So,

the limiting reactant is AlCl₃ and the excess reactant is NaOH.