Chemistry 122 i

Question

Calculate the mass of lead chloride that could be made when 50. mL of 0.800 M lead nitrate reacts with excess sodium chloride according to the following reaction:

Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + NaNO3(aq)

Answer 1

Good morning Bailey! I hope my answer is able to assist your academic endeavors!

The first thing you want to do whenever looking at these problems is to establish that your reaction is completely balanced before doing any number crunching. Even if you know how to solve the problem mathematically it will not mean anything if your reaction is not balanced.

Your reaction with lead nitrate in your question is not balanced on the products side! To fix this, we just have to add a two coefficient in front of NaNO_{3 (aq)}.

Now that we have balanced the reaction, lets move onto the math. The question says that you calculate the amount of lead chloride that can be produced with a given amount of reactant present. This problem is asking of your understanding of both stoichiometry and molarity. First lets focus on the molarity!

We want to establish in moles how much of the reactant is given to us. This will help us with the stoichiometry part. The question states that you start with 50 mL of 0.8 M Lead Nitrate, so by using the definition of molarity, we can calculate the total moles of Lead nitrate.

moles = Molarity x = 0.8 M x = 0.04 moles of Lead Nitrate

Volume (L) 0.05 L

This is where we apply the use of the formula and stoichiometry. We can use the balanced reaction as a ratio in our stoichiometric equation. Since the Lead Nitrate and Lead Chloride are in a 1 to 1 ratio, it does not heavily influence our final answer but it is good to include it in the stoichiometry.

0.04 Moles of Lead Nitrate X 1 Mole of Lead Chloride X 278.1 g of Lead Chloride = 11.124 g PbCl₂

1 Mole of Lead Nitrate 1 Mole of Lead Chloride

I hope my answer is helpful to you!

Answer 2

Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq) ... balanced equation

Use the mole ratio (stoichiometry) of PbCl₂ to Pb(NO₃)₂ to find the answer.

mols Pb(NO₃)₂ used = 50. ml x 1 L /1000 ml x 0.800 mol/L = 0.040 mols

mols PbCl₂ that can be made = 0.040 mols Pb(NO₃)₂ x 1 mol PbCl₂ / 1 mol Pb(NO₃)₂ = 0.04 mol PbCl₂

grams PbCl₂ that can be made = 0.040 mols PbCl₂ x 278 g /mol = 11 g PbCl₂ (2 sig. figs.)