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
