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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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