SOLUTION —
[In order to derive the formula of oxide of nitrogen it is worthwhile to know the volume of oxide of nitrogen formed when a given volume of each of nitrogen and oxygen reacts.]
We proceed in the following way :
Total vol. of oxide of nitrogen + H2 = 28 mL
Vol. of H2 added $=(28-10) mL \quad=18 mL$
From (c) and (d), vol. of $O _{2}$ added $\quad=(27-18) mL =9 mL$
and from (d) and (e), contraction in volume =(27-15)=12 mL.
We must understand that oxide of nitrogen reacts with H2 to form water and N2 gas.
From (c) the volume $18 mL$ must be a mixture of N2 gas and unused hydrogen because oxygen has been added to this volume only to react with unused H2 gas.
Contraction in volume after second explosion =(27-15) mL =12 mL and this contraction is certainly due to formation of water.
Hence vol. of H2 reacted $=12 \times \frac{2}{3} mL =8 mL$
and vol. of O2 reacted $=12 \times \frac{1}{3} mL =4 mL$
From $9 mL$ of $O _{2}$ gas, only $4 mL$ has reacted. Hence, the volume of unreacted $O _{2}=$ $(9-4) mL =5 mL$ and this $5 mL$ of $O _{2}$ must be present in $15 mL$ of gas obtained after second explosion. Hence, the other gas must be N2 gas.
$\therefore$ Vol. of N2 gas formed =(15-5) mL =10 mL.
From analysis of above data we see that $\delta mL$ of H2 must be present in the gas in (c). Thus volume of H2 required for complete reaction with 10mL of oxide of nitrogen = (18-8) mL =10 mL.
We then conclude that $10 mL$ of oxide of nitrogen react completely with $10 mL$ of hydrogen to form 10 mL of N2, and 10 mL of H2 must require 5mL of O2 for complete reaction. That is.
10mL of N2 and 5mL of O2 react to form $10 mL$ of oxide of nitrogen or, $2 mL$ of $N_{2}$ and $1 mL$ of $O_{2}$ react to form $2 mL$ of oxide of nitrogen or, 2 molecules of $N _{2}+1$ molecule of $O _{2}=2$ molecules of oxide of nitrogen or, 1 molecule of $N_{2}+\frac{1}{2}$ molecule of $O_{2}=1$ molecule of oxide of nitrogen or, 2 atoms of $N_{2}+1$ atom of $O_{2}=1$ molecule of oxide of nitrogen Therefore, formula of oxide of nitrogen = N2O.
