Question

If 35.6 L of ammonia gas and 40.5 L of oxygen gas (both at 25.0̊C and 1.100 atm) react according to this equation, what volume (in L) of NO (at 25.0̊C and 1.100 atm) will be produced?

Answer 1

Balanced equation:   4NH₃ + 5O₂ ===> 4NO + 6H₂O

At STP, one mole of any gas has a volume of 22.4 L.

35.6 L NH₃ x (1 mole NH₃ / 22.4 L NH₃) = 1.59 moles NH₃

40.5 L O₂ x (1 mole O₂ / 22.4 L O₂) = 1.81 moles O₂

Now the question is: Which one of the reactants (NH₃ or O₂) will be used up first?

According to the balanced equation above, it takes 5 moles of O₂ to react with every 4 moles of NH₃.

1.59 moles NH₃ x (5 moles O₂ / 4 moles NH₃) = 1.99 moles O₂.

But we have only 1.81 moles of O₂, not enough to react with all the NH₃. So O₂ will be used up first.
How much NO will it make?
According to the balanced equation, 5 moles of O₂ will produce 4 moles of NO.

1.81 moles O₂ x (4 moles NO / 5 moles O₂) = 1.448 mole of NO is produced

Therefore 1.448 x 22.4 = 32.4 L of NO is produced