2KI(aq) + Pb(NO 3 ) 2 (aq) -->PbI 2 (s) + 2KNO 3 (aq) 1.Calculate the mass in g

Question

2KI(aq) + Pb(NO3)2(aq) -->PbI2(s) + 2KNO3(aq) 1.Calculate the mass in g of the PbI2(s)(yellowprecipitate) that is formed after reacting with 1mL of 0.1M KI solution + 1mL of 0.06MPb(NO3)2 solution and then, add another 1mLof 0.1M KI solution. 2. what is the limiting reagent? 2KI(aq) + Pb(NO3)2(aq) -->PbI2(s) + 2KNO3(aq) 1.Calculate the mass in g of the PbI2(s)(yellowprecipitate) that is formed after reacting with 1mL of 0.1M KI solution + 1mL of 0.06MPb(NO3)2 solution and then, add another 1mLof 0.1M KI solution. 1.Calculate the mass in g of the PbI2(s)(yellowprecipitate) that is formed after 2. what is the limiting reagent?

Explanation / Answer

Total KI solution added = 2mL n(KI) = cv = 0.1 x 0.002 = 0.0002mol n(Pb(NO3)2 ) = cv = 0.06 x 0.001 =0.00006mol Stoichiometry ratio = n(KI)/n(Pb(NO3)2 ) =2/1 = 2 Actual ratio = 0.0002/0.00006 = 3.33 AR > SR So, Pb(NO3)2 is the limiting reagent.(ANSWER TO QUESTION 2) n(PbI2) = 1/1 xn(Pb(NO3)2) = 0.00006mol m(PbI2) = nM = 0.00006 x 461 =0.02766g (ANSWER TO QUESTION 1) Hope this helps!

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