Answer:
19.45g of Cu are expect to be produced
Step-by-step explanation:
Based on the reaction:
2Ni + 3CuCl2 → 3Cu + 2NiCl3
2 moles of Ni react with 3 moles of CuCl2 to produce 3 moles of Cu
To solve this question we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of Cu produced and the mass as follows:
Moles Ni -Molar mass: 58.6934g/mol-
12.00g * (1mol / 58.6934g) = 0.204 moles Ni
Moles CuCl2 -Molar mass: 134.45g/mol-
42.00g * (1mol / 134.45g) = 0.312 moles CuCl2
For a complete reaction of 0.204 moles of Ni are required:
0.204 moles Ni * (3 moles CuCl2 / 2 moles Ni) = 0.306 moles of CuCl2
As there are 0.312 moles, CuCl2 is the excess reactant and Ni is limiting reactant.
Moles Cu:
0.204 moles Ni * (3moles Cu / 2 moles Ni) = 0.306 moles of Cu are produced
Mass Cu:
0.306 moles of Cu * (63.546g / mol) =
19.45g of Cu are expect to be produced