According to the following reaction, what mass of PbCl_2 can form from 235 mL of

Question

According to the following reaction, what mass of PbCl_2 can form from 235 mL of 0 110 M KCl solution? Assume that there is excess Pb(NO_3)2. 2 KCl(aq) + Pb(NO_3)_2(aq) rightarrow PbCl_2(s) + 2 KNO_3(aq) 3 59 g 7.19 g 1 80 g 1.30 g 5.94 g Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of K_2S and Fe(NO_3)_2 are mixed. K*(aq) + NO_3 (aq) rightarrow KNO_3(s) Fe^2+(aq) + S^2-(aq) + 2 K^+(aq) + 2 NO_3 (aq) rightarrow FeS(s) + 2 K*(aq) + 2 NO_3^- (aq) Fe^2+*(aq) + S^2-(aq) rightarrow FeS(s) Fe^2+(aq) + S^2-(aq) + 2 K^+(aq) + 2 NO_3 (aq) rightarrow Fe^2*(aq) + S2 (aq) + 2 KNO_3(s) No reaction occurs.

Explanation / Answer

15)

mol of KCl reacted = M(KCl)*V(KCl)
= 0.110 M * 235 mL
= 25.85 mmol
= 0.02585 mol

from given equation, 2 mol of KCl forms 1 mol of PbCl2

so, mol of PbCl2 formed = (1/2)*mol of KCl
= (1/2)*0.02585
= 0.012925 mol

molar mass of PbCl2 = 278.1 g/mol

mass of PbCl2 = molar mass * number of mol
= 278.1 g/mol * 0.012925 mol
= 3.59 g

Answer: 3.59 g

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