What mass of lead (II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0.250 M solution of lead(II) nitrate. a. 13.9 g
b. 27.8 g
c. 3.48 g
d. 6.95 g
e. 1.39 g

Option d. 6.95 g

Explanation:

First of all, we state the reaction:

2NaCl + Pb(NO₃)₂ → PbCl₂ +  2NaNO₃

We determine the moles of each reactant, to state the limiting

Firstly we convert volume frm mL to L

0.200 L . 0.250M = 0.05 moles of NaCl

0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂

Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.

For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl

For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles

(We only have, 0.05 moles of NaCl)

Stoichiometry to the formed product is 2:1

From 2 moles of NaCl I produce 1 mol of PbCl₂

From 0.05 moles I would produce, the half → 0.025 moles

Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g

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