3) In question 1, we find the following balanced equation:
2 Al + 3 CuCl2.2H2O ---> 3 Cu + 2 AlCl3 + 6 H2O
Step 1 - Let's find out how many moles of Al and CuCl2 reacts
For this, we will use the following formula: mole = mass/molar mass
Molar mass of Al: 26.981539 g/mol
Molar mass of CuCl2.2H2O: 147.01 g/mol
Al:
mass = 0.5 g
molar mass = 26.981539 g/mol
mole = 0.5/26.981539
mole = 0.018531189
CuCl2.2H2O:
mass: 3.5 g
Molar mass: 147.01 g/mol
mole = 0.023807904
Step 2 - Let's see which one is the limiting reactant.
2 moles Al --- 3 moles CuCl2.2H2O
0.018531189 moles Al --- x
x = 0.02779678
2 moles Al --- 3 moles CuCl2.2H2O
x moles Al --- 0.023807904 moles CuCl2.2H2O
x = 0.015871936
CuCl2.2H2O is the limiting reactant. It means that 0.023807904 moles CuCl2.2H2O react with 0.015871936 moles of Al and produce 0.023807904 moles of Cu.
Step 3 - Let's transform 0.023807904 moles of Cu into grams. For this, we use the following formula: mass = mole x molar mass
molar mass of Cu = 63.546
mole = 0.023807904 moles
mass = 0.023807904 x 63.546
mass = 1.5129 g
Answer: The theoretical yield of copper produced is 1.5129 g