Mass to Mass Stoichiometry/ Percent Yield (January 26 ' 10)

Converting number of moles into grams.

Mass to Mass examples:
Lead (IV) nitrate reactions with 5.0 g of Potassium iodide. How many grams of Lead (IV) Nitrate are required?
Pb(NO3) + 4 KI -----> 4 KO3 + PbI4
5.0 g x 1 mol KI/165.9 g x 1 Pb(NO3)4/4KI x 455.2/ 1 mol Pb(NO3)4 = 3.4 g

if a 100 mL solution of 2.0 M H2SO4 is neutralized by sodium hydroxide what mass of water is produced?
H2(SO4) + NaOH -----> HOH + Na2(SO4)
2.0 mol/L x .1 L = .200 mol H2(SO4) x 2 HOH/1 H2(SO4) x 18.0 g/mol = 7.2 g H2O


Percent Yield
The theoretical yield of a reaction is the expected (calculated) amount
The experimental amount is the actual yield
% yield = actual/theoretical x 100

Example:
he production of urea CO(NH2) is given by: 2NH3 + Co2 -----> CO(NH2) + H2O. If 47.7 g of urea are produced when 1 mole of CO2 reacts, find the actual yield, theoretical yield and percent yield.

We know that the actual yield is 46.7 g.
The theoretical yield: 1 mol CO2 x 1 CO(NH2)2/1 CO2 x 60.1 g/ 1 mol = 60.1 g

Percent Yield = 47.7 g/60.1 g x 100 = 79.4%