Chemistry question

Question

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

Answer 1

Step 1 - Write the raw chemical reaction equation (just balance the charges on the individual components but don't try to balance the overall equation).

Ni + Pb(C₂H₃O₂)₄ → Ni(C₂H₃O₂)₂ + Pb

(Notice that since acetate has a -1 charge, there are 4 required on the reactant side because it's Lead (IV) and 2 that are required on the product side because it's Ni (II) )

Step 2 - Balance the equation

2Ni + Pb(C₂H₃O₂)₄ → 2Ni(C₂H₃O₂)₂ + Pb

Step 3 - Convert 275g of Pb(C₂H₃O₂)₄ into moles

To do that, calculate the molar mass of Pb(C₂H₃O₂)₄ by adding all the pieces (PB: 1 x 207.2, C: 8 x 12.011, H: 12 x 1.008, O: 8 x 15.999) to get 443.376 g/mol. Then take the 275g and divide by the 443.376 g/mol to get 0.62024 moles of Pb(C₂H₃O₂)₄

Step 4 - Calculate moles of Ni(C₂H₃O₂)₂

Using the molar ratios in the reaction equation, for every 1 mole of Pb(C₂H₃O₂)₄ reacted, there are 2 moles of Ni(C₂H₃O₂)₂ produced. So, multiply the number of moles of Pb(C₂H₃O₂)₄ by 2 to get moles of Ni(C₂H₃O₂)₂: 0.62024 x 2 = 1.24048 moles but since there are only 3 significant figures given, round to 1.24 moles of Nickel(II) Acetate