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

yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet

the answer to this question is 3.