Consider the following reaction. N2 + 3 H2 → 2 NH3 . 28.0 grams of N2 and 5.04 grams of H2 are reacted, producing 17.8 grams of NH3. What is the percent yield?

Answer 1

1) Find the limiting reactant:

moles N2 = x

moles H2 = y

Divide x by 1 and y by 3 and whichever is smaller is the limiting reactant

Theoretical yield = moles limiting reactant x moles 2 mol NH3/mol LR x molar mass NH3 = g NH3

Percent yield = actual/theoretical (x100%) = 17.8 g / theoretical (x100%) = answer