Answer:
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
I hope it helps!
