If CuCl2 is dihydrate, the equation of the reaction should be:
2 Al + 3 CuCl2*2H2O → 3 Cu + 2 AlCl3 + 6 H2O
So the answer for number one is: 2 Al + 3 CuCl2*2H2O → 3 Cu + 2 AlCl3 + 6 H2O
Then, we should do the calculation to find out if question 2 is right:
Step 1 - Transform grams of Al and CuCl2*2H2O into moles using the following formula: mole = mass/molar mass
molar mass of Al = 26.981539
molar mass of CuCl2*2H2O = 170.48256 g/mol
mass of Al= 0.5/26.981539 = 0.018531189 moles
mass of CuCl2*2H2O = 3.5/170.48256 = 0.020529959
Step 2 - Use the equation proportion to find the limiting and excess reactant.
2 moles Al --- 3 moles CuCl2*2H2O
0.018531189 moles Al --- x moles CuCl2*2H2O
x = 0.027796788 moles of CuCl2*2H2O
2 moles Al --- 3 moles CuCl2*2H2O
x moles Al --- 0.020529959 moles CuCl2*2H2O
x = 0.013686639 moles of Al
As we can see, CuCl2*2H2O is the limiting reactant.
