Question

Consider the following reaction: 2 A1 + 3 CuCl2 → 3 Cu + 2 AICI: Assuming 23.35 g of Al are reacted with 44.89 g of copper Il chloride, how much of the excess reactant is left behind (unused)? | ban Identify the limiting reactant: OAL O CuCl2

Answer 1

the equation is as follows

2 Al + 3 CuCl2 ---------------------> 3 Cu + 2 AlCl3

no of moles = mass / molar mass

for Al = 23.35 g / 26.98 g/mol => 0.86545 moles

for CuCl2 = 44.89 g / 134.45 g/mol => 0.334 moles

=> 3 mol CuCl2 -----------------------> 2 mol Al

0.334 mol ------------------------------>?

=> 0.334 * 2 / 3

=> 0.2225 mol of Al required

so limiting reactant is CuCl2

excess reactant is Al

excess reactant moles => 0.86545 - 0.2225 => 0.64287 moles

mass of excess reactant unused => 0.64287 mol * 26.98 g/mol => 17.345 g

answers => 17.345 g

excess reactant is Al