Amateur chemists make nitrous oxide at home using this (unbalanced) reaction:

How many grams of NH_{4}NO_{3} are required to produce 134.4 L of mixed gas if products are at STP?

(Do NOT heat the ammonium nitrate above 240°C or it may detonate)

- 160
- 240
- 960
- 1440

**Explanation**

This solution has 3 steps: balance the equation, determine the required number of moles of product and reactant, and calculate the mass of reactant necessary. The reaction as written is unbalanced. In the balanced equation, 2 moles of H_{2}O and 1 mole of N_{2}O are made for each 1 mole of NH_{4}NO_{3}. 3 moles of product gas result from each mole of reactant. 22.4 L of gas corresponds to 1 mole of gas at STP, and the specified 134.4 L indicates 6 moles of product gases. 6 moles of mixed product gas requires 2 moles of reactant. NH_{4}NO_{3} weighs 80 g/mole, so 160 g are required.

A) 160, correct. 2 moles of NH_{4}NO_{3} yields 4 moles of H_{2}O and 2 moles of N_{2}O, for 6 moles of mixed product gas occupying 134.4 L at STP.

B) 240, incorrect, This answer results from using the unbalanced equation as written.

C) 960, incorrect, This answer results from using the unbalanced equation and calculating the mass of 12 moles NH_{4}NO_{3} for 6 moles N_{2}O product rather than mixed gas.

D) 1440, incorrect, This answer results from using the correct balanced equation and calculating the mass of 18 moles of NH_{4}NO_{3} for 6 moles of N_{2}O product rather than mixed gas.

Just a minor correction- the answer explanation is referring to ammonium ammonium? (NH4NH3) rather than the correct ammonium nitrate. Doesn’t change the answer but it definitely confused me for a minute.

Hi Hamad, thank you very much for pointing this out! Our mistake. The explanation has now been corrected to refer to ammonium nitrate, as we intended.