nickel metal is put into lead (iv) acetate solution and produces nickel (ii) acetate and solid lead. if 275 g of lead (iv) acetate reacts with excess nickle. how many moles of nickle (ii) acetate will be produced

1.24 mol.

Explanation:

Ni metal reacts with Pb (CH₃COO) ₄ according to the balanced equation:

2Ni (s) + Pb (CH₃COO) ₄ → 2Ni (CH₃COO) ₂ + Pb (s),

It is clear that 2 mole of Ni metal reacts with 1 mole of Pb (CH₃COO) ₄ to produce 2 mole of Ni (CH₃COO) ₂, and 1 mole of Pb.

Firstly, we need to calculate the no. of moles of 275.0 g of Pb (CH₃COO) ₄:

no. of moles of Pb (CH₃COO) ₂ = mass/molar mass = (275.0 g) / (443.38 g/mol) = 0.62 mol.

using cross multiplication:

1.0 mol of Pb (CH₃COO) ₄ produces → 2 mol of Ni (CH₃COO) ₂, from the stichiometry.

0.62 mol of Pb (CH₃COO) ₄ produces →? mol of Ni (CH₃COO) ₂.

∴ The no. of moles of Ni (CH₃COO) ₂ are produced = (0.62 mol) (2.0 mol) / (1.0 mol) = 1.24 mol.