The balanced chemical equation is Pb(NO3)2(aq) + 2KCl(aq) ---> PbCl2(s) + 2KNO3(aq)
Molar mass(g/mole) 331.2 74.5 278.1
From the balanced equation ,
1 mole = 331.2 g of Pb(NO3)2 reacts with 2 mole = 2*74.5 = 149 g of KCl
7.42 g Pb(NO3)2 reacts with M g of KCl
M = ( 7.42g*149g)/331.2g
= 3.34 g of KCl
So 6.44-3.34 g = 3.10 g of KCl left unreacted so KCl is the excess reactant.
Since all the mass of Pb(NO3)2 reacted completely it is the limiting reactant.
Again from the balanced equation,
1 mole = 331.2 g of Pb(NO3)2 produces 1 mole = 278.1g of PbCl2
7.42 g of Pb(NO3)2 produces N g of PbCl2
N = ( 7.42g * 278.1g) / 331.2g
= 6.23 g of PbCl2
So the theoretical yield of PbCl2 is 6.23g
That means for 100% yield the mass of precipitate produced is 6.23g
For 89.3% the mass of precipitate produced is (89.3% *6.23g) / 100% = 5.56g
An aqueous solution containing 7.42 g of lead(II) nitrate is added to an aqueous solution containing...
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