1.15 g H2 is allowed to react with 10.0 g N2, producing 2.24 g NH3.
What is the theoretical yield in grams for this reaction under the given conditions?
What is the percent yield for this reaction under the given conditions?
N2 + 3H2 ---> 2NH3
moles of H2 = g/molar mass = 1.15 g/2.016 g/mol = 0.570 mols
moles of N2 = 10.0 g/28 g/mol = 0.357 mols
If all of N2 reacts we would need = 3 x 0.357 = 1.071 mols H2
If all of H2 reacts we would need = 0.570/3 = 0.190 mols N2
Since moles of H2 are less then required, H2 is the limiting reagent
Theoretical yield of NH3 = 0.570 mol x 2 x 17.031 g/mol/3 = 6.47 g
Percent yield = (experimental yield/theoretical yield) x 100
= (2.24/6.47) x 100
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