પ્રક્રિયકના પ્રતિ ગ્રામ માટે , નીચેનામાંથી કઈ ઉષ્મિય વિઘટન પ્રક્રિયામાં મહત્તમ પ્રમાણમાં $N_2$ વાયુ ઉત્પન્ન થાય છે?

Question

JEE MAIN 2018, Diffcult

Download our app for free and get started

Solution

c
$(a)$ Molar mass of $Ba(N_3)_2(s) = 221\,g/mol$

$1$ mole of $Ba(N_3)_2(s)$ will give $3$ moles of $N_2$

hence $\frac{{1\,g}}{{221\,g/mol}}$ moles of $Ba(N_3)_2(s)$ will give $3 \times \frac{1}{{221}} = 0.014$ moles of $N_2$

$(b)$ Molar mass of ${(N{H_4})_2}C{r_2}{O_7} = 252\,g/mol.\,$

$1$ mole of ${(N{H_4})_2}C{r_2}{O_7}$ will give $1$ mole of $N_2$

hence $\frac{{1\,g}}{{252\,g/mol}}$ moles of ${(N{H_4})_2}C{r_2}{O_7}$ will give $1 \times \frac{1}{{252}} = 0.0039$ moles of $N_2$

$(c)$ Molar mass of $NH_3 = 17\,g/mol.$

$2$ mole of $NH_3$ will give $1$ mole of $N_2$

hence $\frac{{1\,g}}{{17\,g/mol}}$ moles of $NH_3$ will give $\frac{1}{{2 \times 17}} = 0.0297$ moles of $N_2.$

$(d)$ Molar mass of $NH_4NO_3 = 80\,g/mol.$

$1$ mole of $NH_4NO_3$ will give $1$ mole of $N_2$

hence $\frac{{1\,g}}{{80\,g/mol}}$ moles $NH_4NO_3$ will give $1 \times \frac{1}{{80}} = 0.0125$ moles of $N_2$

Hence Thermal decomposition of $NH_3$ will produce maximum amount of $N_2$

Download our app
and get started for free