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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

Respuesta :

superman1987

Answer:

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.

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