For per gram of reactant,the maximum quantity | vedclass

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(C) To find the maximum quantity of $N_2$ produced per gram of reactant,we calculate the moles of $N_2$ produced per gram of each reactant:$(a)$ Molar

Vedclass · Accepted Answer

$2NH_{3(g)} 	o N_{2(g)} + 3H_{2(g)}$

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(C) To find the maximum quantity of $N_2$ produced per gram of reactant,we calculate the moles of $N_2$ produced per gram of each reactant:$(a)$ Molar mass of $Ba(N_3)_2 = 137 + 6 	imes 14 = 221 \ g/mol$. $1 \ mol$ of $Ba(N_3)_2$ produces $3 \ mol$ of $N_2$. $N_2$ produced per gram $= \frac{3}{221} \approx 0.0136 \ mol$.$(b)$ Molar mass of $(NH_4)_2Cr_2O_7 = 2(14+4) + 2(52) + 7(16) = 36 + 104 + 112 = 252 \ g/mol$. $1 \ mol$ of $(NH_4)_2Cr_2O_7$ produces $1 \ mol$ of $N_2$. $N_2$ produced per gram $= \frac{1}{252} \approx 0.0040 \ mol$.$(c)$ Molar mass of $NH_3 = 14 + 3 = 17 \ g/mol$. $2 \ mol$ of $NH_3$ produces $1 \ mol$ of $N_2$. $N_2$ produced per gram $= \frac{1}{2 	imes 17} = \frac{1}{34} \approx 0.0294 \ mol$.$(d)$ Molar mass of $NH_4NO_3 = 14 + 4 + 14 + 48 = 80 \ g/mol$. $2 \ mol$ of $NH_4NO_3$ produces $2 \ mol$ of $N_2$ (i.e.,$1 \ mol$ of $NH_4NO_3$ produces $1 \ mol$ of $N_2$). $N_2$ produced per gram $= \frac{1}{80} = 0.0125 \ mol$.Comparing the values,$0.0294 > 0.0136 > 0.0125 > 0.0040$. Thus,$NH_3$ produces the maximum amount of $N_2$.

For per gram of reactant,the maximum quantity of $N_2$ gas is produced in which of the following thermal decomposition reactions? (Given: Atomic wt. : $Cr = 52 \ u, Ba = 137 \ u$).

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