Basic chemistry

Question

What volume of NH3 will appear (stp) after reaction of 13,2 grams of NH42SO4 and 5,6 grams of KOH?

Answer 1

Write the balanced equation...

(NH₄)₂SO₄ + 2KOH ==> K₂SO₄ + 2NH₄OH ==> 2H₂O + 2NH₃

Next, find out which reactant is limiting ...

moles (NH₄)₂SO₄ = 13.2 g x 1 mol/132 g = 0.100 moles

moles KOH = 5.6 g x 1 mol/56.1 g = 0.100 moles

Since the balanced equation shows it takes rwice as many moles KOH as NH₄)₂SO₄ , the KOH is limiting and will determine the maximum amount of NH₃ that can be formed.

Finally, find the moles of NH₃ and then convert to volume at STP ...

moles NH₃ = 0.1 mol KOH x 2 mol NH₃/2 mol KOH = 0.1 mol NH₃

Volume @STP = 0.100 mol x 22.4 L/mol = 2.24 liters NH₃