Chemistry Help!?

Question

A recation combines 113.484 g of lead (II) nitrate with 45.010 g of sodium hydroxide.

-If the actual yield of lead (II) hydroxide were 80.02 g, what was the percent yield

Answer 1

1 Pb(NO3)2  & 2 NaOH --> 1 Pb(OH)2  & 2 NaNO3

using molar masses:

1 mole Pb(NO3)2 = 331.2 grams/ 1 mole

2 moles NaOH = 80.00 grams / 2 moles

1 mole Pb(OH)2 = 241.2 grams/1 mole

we need a ratio of 331.2 g Pb(NO3)2 / 80 g NaOH = 4.14

they provided a ratio of 113.484g / 45.010 = 2.52

so they have not provided enough Pb(NO3)2,

Pb(NO3)2 is your limiting reagent:

? g Pb(OH)2 can be produced from:

113.484 g Pb(NO3)2 @  241.2 g Pb(OH)2 / 331.2 g Pb(NO3)2 = 82.646 grams of Pb(OH)2

% yield :

80.02 g / 82.65 g times 100 =

your answer: 96.82%