Question

# In an experiment, $10c{{m}^{3}}$ of an organic compound, J, in the gaseous state is reacted with an excess of oxygen. Steam, $20c{{m}^{3}}$ of carbon dioxide and $5c{{m}^{3}}$ of nitrogen are the products.All gas volumes were measured at the same temperature and pressure.What could be the identity of J?1) ${{C}_{2}}{{H}_{6}}{{N}_{2}}$ 2) ${{C}_{2}}{{H}_{3}}N$ 3) ${{C}_{2}}{{H}_{7}}N$ (a)- 1, 2, and 3 are correct(b)- 1 and 2 only are correct(c)- 2 and 3 only are correct(d)- 1 only is correct

Hint: The volume of the ideal gas is taken at STP (standard temperature and pressure). Its value is 22.7 L. The number of moles is equal to the given volume divided by the volume of an ideal gas at STP. With the schematic equation, we calculate the number of carbon and nitrogen atoms.

1 mol of an ideal gas has the volume at STP = 22.4 L.
Hence, ${{V}_{m}}=\text{ }22.4L$
So, the number of moles is equal to the given volume upon volume of 1 mol gas.
We have to find the number of atoms of carbon and nitrogen in C H N.
So, we can account x to C, y to H, and z to N.
$({{C}_{x}}{{H}_{y}}{{N}_{z}})$
Therefore, number of moles of the compounds will be:
$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}{{{V}_{m}}}=\dfrac{10c{{m}^{3}}}{{{V}_{m}}}$
$n({{N}_{2}})=\dfrac{V({{N}_{2}})}{{{V}_{m}}}=\dfrac{5c{{m}^{3}}}{{{V}_{m}}}$
$n(C{{O}_{2}})=\dfrac{V(C{{O}_{2}})}{{{V}_{m}}}=\dfrac{20c{{m}^{3}}}{{{V}_{m}}}$
According to the question, the schematic equation of the chemical reaction will be:
${{C}_{x}}{{H}_{y}}{{N}_{z}}\text{ }+\text{ }(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O+\dfrac{z}{2}{{N}_{2}}$
So, by formulating with the help of the above reaction we can find the value of x and y.
We can write,
$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{n(C{{O}_{2}})}{x}=\dfrac{2\text{ x }n({{N}_{2}})}{z}$
For, x
$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{n(C{{O}_{2}})}{x}$
$x=\dfrac{n(C{{O}_{2}})}{n({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\frac{V(C{{O}_{2}})}{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{20c{{m}^{3}}}{10c{{m}^{3}}}=2$
So, the value of x is 2
For, z
$\dfrac{z}{2}=\dfrac{n({{N}_{2}})}{n({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{V({{N}_{2}})}{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{5c{{m}^{3}}}{10c{{m}^{3}}}=\dfrac{1}{2}$
$z=\dfrac{1}{2}\text{ x 2 = 1}$
So, the value of z is 1.
So, in the J compound, the carbon atoms are 2, and the number of nitrogen atoms is 1.
Hence, the correct option is (c)- 2, and 3 only are correct.
So, the correct answer is “Option C”.

Note: We have directly put the value of the volume of a given compound because the volume of an ideal gas is the same for every compound, hence it gets canceled. The schematic equation of the chemical reaction should be correct.