ES Additional Exercise 9.45 4When a solution of lead(II) nitrate is mixed with a
ID: 706004 • Letter: E
Question
ES Additional Exercise 9.45 4When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balanced equation for the reaction, including the states of all the substances. .45 2 Edit udy SHOW HINT (b) When 19.94 g of lead(I) nitrate are mixed with 13.30 g of sodium chromate, what is the percent yield of the solid if 6.93 g are recovered? Percent yield: the tolerance is +/-9% SHOW HINT SHOW HINT LINK TO TEXT Prvacy Policy1 n Wiley & 5ons Ins. All Rights Reserved, A Division of JohnExplanation / Answer
Pb(NO3)2 (aq) + Na2CrO4 (aq) ---> PbCrO4 (s) + 2NaNO3 (aq)
is the reaction
b)Pb(NO3)2 moles = mass / ( molar mass of Pb(NO3)2) = 19.94g / ( 331.2g/mol) = 0.06 mol
Na2CrO4 moles = ( 13.3g) / ( 162 g/mol) = 0.082 mol
as per reaction both reactants react in 1:1 ratio. Since Pb(NO3)2 is relatively less it is limiting reagent
PbCrO4 moles formed = Pb(NO3)2 moles reacted = 0.06 mol
PbCrO4 mass = PbCrO4 moles x molar mass of PbCrO4
= 0.06 mol x 323.2g/mol = 19.39 g = theoretical yield
% yield = ( 100 x actual yield / theoretical yield)
= ( 100 x 6.93g / 19.39g) = 35.74 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.