According to the following reaction, what mass of PbCl2 can form from 235 mL of 0.110 M KCl solution? Assume that there is excess Pb(NO3)2.

3,59g of PbCl₂(s) can be formed

Explanation:

Based on the reaction:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

Moles of 235mL ≡0,235L of 0,110M KCl are:

0235L×0,110mol/L = 0,02585 moles of KCl. As 2 moles of KCl produce 1 mole of PbCl₂(s):

0,02585 moles of KCl × (1 mole PbCl₂(s) / 2 mol KCl) = 0,01293 moles of PbCl₂(s)

In grams (Molecular mass of PbCl₂(s) is 278,1g/mol):

0,01293 moles of PbCl₂(s) × (278,1g / 1mol) = 3,59g of PbCl₂(s) can be formed

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