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