Question

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)
NH₄NO₃ $\stackrel{}{\to }$ N₂O + 2H₂O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam?
(Relative molecular mass of ammonium nitrate is 80.)
(iii) Determine the percentage of oxygen in the ammonium nitrate. (O = 16)

Answer 1

(i) According to the balanced chemical equation, for one mole of dinitrogen oxide, two moles of steam is produced.
According to Avogadro's Law, at STP, same number of gaseous molecules occupy equal volumes.
Volume of dinitrogen oxide will be half of the volume occupied by steam.
Volume of dinitrogen oxide = $\frac{8.96}{2}=4.48\mathrm{L}$

(ii) Number of moles in 8.96 litres of steam = $\frac{8.96}{22.4}=0.4\mathrm{mol}$
Number of moles of NH₄NO₃ will be half of the moles of steam formed.
Number of moles of NH₄NO₃ = $\frac{0.4}{2}=0.2\mathrm{mol}$
Mass of NH₄NO₃ in 0.2 moles = 0.2 $\times$ Molecular mass of NH₄NO₃
= 0.2 $\times$ 80 =16 g

(iii) Mass of oxygen in NH₄NO₃ = 48 g
$\mathrm{Percentage}\mathrm{of}\mathrm{O}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{O}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_4{\mathrm{NO}}_3}\times 100=\frac{48}{80}\times 100=60\%$