Limiting reagent problems, basically, are two stoichiometry problems in one.
There is a "little" shorter method but I like the following one.
1. You have the equation.
2. Convert 18.1 g NH3 to moles. moles = grams/molar mass.
3. Convert 90.4 g CuO to moles the same way.
4. Using the coefficients in the balanced equation, convert moles NH3 to moles N2.
5. Same process, convert moles CuO to moles N2.
6. It is more than likely that the answers from the last two steps will not agree which means one of the answers for moles N2 is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
7. Using the value from the last step (the smaller one), convert moles to grams. g = moles x molar mass. This is the theoretical yield.
8. percent yield = (actual yield/theoretical yield)*100 = ??
consider the following balance chemical equation:
2NH3(g)+3CuO(s)=N2(g)+3Cu(s)+3H2O(g)
if 18.1grams of ammonia reacts with 90.4 grams of copper(II) oxide, which is the limitting reagent?? what is the theorectical yield of nitrogen gasd? if 8.52 g of N2 are formed, what is the % yield of nitrogn gas??
1 answer
2 answers