Question

$34gm$ of a mixture containing ${N_2}$ and ${H_2}$ in $1:3$ by mole is partially converted into $N{H_3}$ . Calculate the vapor density of the mixture containing remaining ${N_2},{H_2}$ and $N{H_3}$ after reaction, if it has been found that the $N{H_3}$ form required $0.5moles$ of ${H_3}P{O_4}$ for complete neutralization.
$3N{H_3} + {H_3}P{O_4} \to {(N{H_3})_3}P{O_4}$

Answer 1

Hint: Vapor density is the density of vapor in relation to that of hydrogen. It may be defined as mass of a certain volume of a substance divided by the mass of the same volume of hydrogen. It is also defined as the molecular weight of the given compound with respect to the molecular weight of hydrogen gas. Thus, it is the half of the molecular weight of a given compound.

Complete step by step answer:
The vapor density indicates whether a gas is denser (greater than one) or less dense (less than one) than the air. Mathematically, vapor density can be written as:
$V.D = \dfrac{{{{({M_w})}_{N{H_3}}}}}{{{{({M_w})}_{{H_2}}}}}$
Where, $V.D = $ vapor density of ammonia
${({M_w})_{N{H_3}}} = $ molecular weight of ammonia = $(14 \times 1) + (1 \times 3) = 17g$
${({M_w})_{{H_2}}} = $ molecular weight of hydrogen = $2g$
But, the number of moles of ammonia formed initially = $\dfrac{{34}}{{17}} = 2moles$
The number of moles of ammonia used up on reaction with phosphoric acid = $(2 - (0.5 \times 3))moles = 1.5moles$
Thus, the vapor density of the mixture of hydrogen and nitrogen gas = $\dfrac{{{N_2} + 3{H_2}}}{2} = \dfrac{{28 + 6}}{2} = \dfrac{{34}}{2} = 17$ .


Note:
 The density has implications for container storage and personnel safety, which means if a container can release a dense gas, its vapor could sink and, if flammable, collect until it is at a concentration sufficient for ignition. Even if not flammable, it could collect in the lower floor or level of a confined space and displace air, possibly presenting an asphyxiation hazard to individuals entering the lower part of that space.