mols Cu initially = grams/atomic mass = about 0.3 but you need a more accurate answer.
mols HNO3 initially = M x L = about 1.3 but see note above.
Convert mols Cu to mols NO2. That's about 0.3 mols Cu x (2 mols NO2/1 mols Cu) = about 0.6 mols.
Convert mols HNO3 to mols NO2. That's 1.3 mols HNO3 x (2 mols NO2/4 mols HNO3) = 0.65 (remember is is approx--you go through this more accurately).
In limiting reagent problems the smaller value is ALWAYS the correct value and the reagent producing that number is the limiting reagent. In this Cu is the limiting reagent and HNO3 is the excess reagent.
0.6 mols NO2 produced. Convert to grams. g = mols x molar mass = ? This is the theoretical yield (TY). The actual yield (AY) is 22.5 g.
Then %yield = (AY/TY)*100 = ?
Post your work if you get stuck.
an 18.9g sample of copper and an 82.0 mL of 16.0mol/L nitric acid are allowed to react. Brown nitrogen dioxide gas is generated.
find the maximum mass of NO2 that could be produced.
if 22.6g of the gas is actually produced, find the percentage yield of the reaction.
Cu + 4HNO3 -> Cu(NO3)2 + 2H2O +2NO2
1 answer
1 answer