Question

# $$60$$ mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If $$38$$ mL of $${N}_{2}$$ was formed, then the volume (in mL) of nitric oxide in the mixture is (all measurements are made at constant $$P$$ and $$T$$) :

Solution

## Let the volume of $$NO\ and\ {N}_{2}O$$ be $$a$$ and $$b$$ mL respectively.Then, $$a+b=60 ...(i)$$$$NO+{H}_{2}\rightarrow \frac{1}{2}{N}_{2}+{H}_{2}O$$    $$a$$ mL                 $$0$$    $$0$$                        $$\displaystyle\frac{a}{2}$$$${N}_{2}O+{H}_{2}\rightarrow {N}_{2}+{H}_{2}O$$  $$b$$ mL                    $$0$$  $$0$$                          $$b$$$$\therefore$$ Total $${N}_{2}$$ formed$$=(\dfrac{a}{2})+b=38$$        ...(ii)By equations (i) and (ii), $$a=44$$ mL, $$b=16$$ mL.So, volume of nitric oxide $$NO = 44$$ mL.Chemistry

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