Looks like the real equation is
"2Al + 3CuSO_4 => 3Cu + Al_2(SO_4)_3"
CuS -- is not copper (II) sulfate and is not soluble in water.
First find the amounts of reagents given to find out which reagent is in excess:
"\\nu (Al) = 35\/27 = 1.3 (mol)"
"\\nu (CuSO_4) = 35\/160 = 0.21875 (mol)"
As 2 mol of Al require 3 mol of CuSO4, then 1.3 mol of Al react with 1.95 mol of CuSO4, but we have less. So we calculate the mass of copper using CuSO4.
1 mol of CuSO4 contains 1 mol of Cu, then 0.21875 mol of CuSO4 contains 0.21875 mol of Cu.
"m = M * \\nu"
"m(Cu) = 64 * 0.21875 = 14 (g)"
Answer: 14 g of copper metal we can produce if we begin the reaction with 35.00 g of solid aluminum metal and 35.00 g of aqueous copper (II) sulfate
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