Question

4NH3 + 5O2 4NO +6H2O

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. How many grams of NO are formed?


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Answer 1

@unknownshadow
4NH3 + 5O2 --> 4NO + 6H2O
3.25g NH3 / 17g/mole = 0.191moles NH3
3.5g O2 / 32g/mole = 0.109moles O2
0.191moles NH3 x (5 O2 / 4 NH3) = 0.239moles O2 required
we only have 0.109moles O2, O2 = limiting reactant

0.109moles O2 x (4NO / 5 O2) x 30g/mole = 2.62g NO produced
0.109moles O2 requires 4/5 x the amount of NH3 or 0.087moles NH3
0.191moles NH3 - 0.087moles NH3 used = 0.104moles left over
0.104moles x 17g/mole = 1.76g NH3 excess

Ref: https://in.answers.yahoo.com/question/index?qid=20120814193931AAErIeF