Please help me with chemistry problem!?!?

J

John S

New member
100.0g of Cu(NO3)2 is added to 100.0g of NaOH, according to balanced equation shown: Cu(NO3)2 + NaOH --> Cu(OH)2 + 2 NaNO3

1. Find the limiting reactant
2. How many grams of Cu(OH)2 can be produced?
3. If only 40.0g of Cu(OH)2 are produced, what is the percent yield?
 
Cu(NO3)2 + 2 NaOH = Cu(OH)2 + 2 NaNO3
moles Cu(NO3)2 = 100.0 g/ 187.56 g/mol=0.533
moles NaOH = 100.0 g /39.9968 g/mol= 2.50 ( in excess)

1
Cu(NO3)2 is the limiting reactant
2.50/2 = 1.25 moles are needed

2
moles Cu(OH)2 produced = 0.533
mass Cu(OH)2 = 0.533 mol x 97.561 g/mol=52.0 g ( theoretical amount)

3
% yield = 40.0 x 100 / 52.0=76.9
 
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