JEE Main 201815 AprilEvening ShiftOnline

JEE Main 2018 · 15 April · Evening Shift · Online · Chemistry · Some Basic Concepts of Chemistry

For per gram of reactant, the maximum quantity of N_2 gas is...

For per gram of reactant, the maximum quantity of

\mathrm{N}_2

gas is produced in which of the following thermal decomposition reactions? (Given: Atomic wt. :

\mathrm{Cr}=52 \mathrm{u}, \mathrm{Ba}=137 \mathrm{u}

Options

A $\mathrm{Ba}$ \left(\mathrm{N}_3\right)_2 $(s)$ \longrightarrow \mathrm{Ba}(\mathrm{C})+3 \mathrm{~N}_2(\mathrm{~g})
B $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7(\mathrm{~s}) \longrightarrow$ $\mathrm{N}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{Cr}_2 \mathrm{O}_3(\mathrm{~s})$
C $2 \mathrm{NH}_3(\mathrm{~g}) \longrightarrow \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$
D $2 \mathrm{NH}_4 \mathrm{NO}_3(\mathrm{~s}) \longrightarrow$ $2 \mathrm{~N}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g})$

C. $2 \mathrm{NH}_3(\mathrm{~g}) \longrightarrow \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$

(a) Molar mass of Ba

\left(\mathrm{N}_3\right)_2(\mathrm{~s})=221 \mathrm{~g} / \mathrm{mol}

1 mole of

\mathrm{Ba}\left(\mathrm{N}_3\right)_2(\mathrm{~s})

will give 3 moles of

\mathrm{N}_2

hence

\frac{1 \mathrm{~g}}{221 \mathrm{~g} / \mathrm{mol}}

moles of

\mathrm{Ba}\left(\mathrm{N}_3\right)_2(\mathrm{~s})

will give

3 \times \frac{1}{221}=0.014

moles of

\mathrm{N}_2

(b) Molar mass of

\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7=252 \mathrm{~g} / \mathrm{mol}

. 1 mole of \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7

will give 1 mole of

\mathrm{N}_2

hence

\frac{1 \mathrm{~g}}{252 \mathrm{~g} / \mathrm{mol}}

moles of \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7

will give

1 \times \frac{1}{252}=0.0039

moles of

\mathrm{N}_2

\mathrm{NH}_3=17 \mathrm{~g} / \mathrm{mol}

. 2 mole of

\mathrm{NH}_3

will give 1 mole of

\mathrm{N}_2

hence

\frac{1 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol}}

moles of

\mathrm{NH}_3

will give

\frac{1}{2 \times 17}=

0.0297

\text{ moles\text{ of } } \mathrm{N}_{2} \text{. } (d) Molar mass of \mathrm{NH}_4 \mathrm{NO}_3=

\mathrm{~g}

\mathrm{mol}. 1 mole

\mathrm{NH}_4 \mathrm{NO}_3 will give 1 mole of \mathrm{N}_2 hence \frac{1 \mathrm{~g}}{80 \mathrm{~g} / \mathrm{mol}} moles \mathrm{NH}_4 \mathrm{NO}_3 will give 1 \times \frac{1}{80}=0.0125 \text{ moles of } \mathrm{N}_{2}

Hence Thermal decomposition of

\mathrm{NH}_3

will produce maximum amount of

\mathrm{N}_2