For the night yet balanced equation, No + CuCl2 —> Cu + NiCl3, if you started with 12.00 g of nickel and 42.00 g of CuCl2, how much Copper would you expect to produce Perdue

19.45g of Cu are expect to be produced

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) =

the ratio of the conjugate base to the acid required to attain the desired ph.