Question

Due in 24 minutes. Due Tue 10/29/2019 Show Intro/Ins Consider the following reaction: 2 Al 3 CuCl2 3 Cu + 2 AICI3 Assuming 65.48 g of Al are reacted with 48.12 g of copper Il chloride, how much of the excess reactant is left behind (unused)? Identify the limiting reactant: CuCl2 6 Al

Answer 1

Ans :

Mol Al = mass / molar mass

= 65.48 g / 26.98 g/mol = 2.427 mol

mol copper II chloride = 48.12 g / 134.452 g/mol = 0.358 mol

2 mol Al needs to react with 3 mol CuCl2

but here amount of CuCl2 is less , so limiting reagent is CuCl2

3 mol CuCl2 reacts with 2 mol Al

so 0.358 mol CuCl2 will react with : ( 0.358 x mol x 2 mol) / 3 mol = 0.238 mol Al

So excess Al = 2.427 mol - 0.238 mol = 2.189 mol

Mass of excess Al left = mol x molar mass

= 2.189 mol x 26.98 g/mol

= 59.05 g