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Question

A student has 7.15 g of zinc powder, 1.30 L of a 3.00 M calcium nitrate solution, and 1.80 L of a 3.40 M lead(II) nitrate solution.

Write the chemical reaction that occurs between the two reacting species. Include physical states.

molecular equation:

What volume of the solution is needed to completely react all 7.15 g of zinc powder? 

volume:

Calculate the molarity of the Zn2+ ions in solution after the reaction. 

concentration:

Answer 1

Strange question. There are actually more than 2 reacting species, so not sure what the question is seeking. For example, there would be a reaction between Zn and Ca(NO₃)₂ as well as between Zn and Pb(NO₃)₂. So there are 3 reacting species. Oh well...here is my answer as best as I can understand the question. Maybe another tutor will have a different perspective.

Zn(s) + Ca(NO₃)₂(aq) ==> Zn(NO₃)₂(aq) + Ca(s)

Zn(s) + Pb(NO₃)₂(aq) ==> Zn(NO₃)₂(aq) + Pb(s)

7.15 g Zn x 1 mol Zn / 65.39 g = 0.109 mols Zn

vol of Ca(NO₃)₂ = 0.109 mol Zn x 1 mol Ca(NO₃)₂/mol Zn x 1L/3.00 mol = 0.0364 L of Ca(NO₃)₂

vol of Pb(NO₃)₂ = 0.109 mol Zn x 1 mol Pb(NO₃)₂/mol Zn x 1 L/3.40 mol = 0.0321 L of Pb(NO₃)₂

Concentration of Zn²⁺ = 0.109 mol Zn²⁺ / 0.0364 L = 2.99 M

Concentration of Zn²⁺ = 0.109 mol Zn²⁺ / 0.0321 L = 3.40 M