When 0.200 grams of al reacts with 15.00 ml of a 0.500 m copper(ii) chloride solution, how many moles of solid cu would be produced? how many grams of solid cu would be produced? show all calculations including the mole-to-mole ratios used in the calculation.

When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
        
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute: 
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025=  0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
 = 0.0075 * 63.546 =0.477 g