JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 22)
For per gram of reactant, the maximum quantity of N2 gas is produced in which of the following thermal decomposition reactions ?
(Given : Atomic wt. - Cr = 52 u, Ba = 137 u)
(Given : Atomic wt. - Cr = 52 u, Ba = 137 u)
(NH4)2Cr2O7(s) $$ \to $$ N2(g) + 4H2O(g) + Cr2O3(s)
2NH4NO3(s) $$ \to $$ 2 N2(g) + 4H2O(g) + O2(g)
Ba(N3)2(s) $$ \to $$ Ba(s) + 3N2(g)
2NH3(g) $$ \to $$ N2(g) + 3H2(g)
Explanation
(a) Molar mass of (NH4)2 Cr2O7 = 252 g/mol.
252g of (NH4)2 Cr2 O7 produce 28g mole of N2
$$\therefore\,\,\,$$ 1 g of (NH4)2 Cr2 O7 Produce = $${{28} \over {252}} = 0.111$$ g N2
(b) Molar mass of NH4 NO3 = 80 g/mol
2 $$ \times $$ 80 g of NH4 NO3 produce 28 $$ \times $$ 2g of N2
$$\therefore\,\,\,$$ 1 g of NH4 NO3 produce = $${{28 \times 2} \over {2 \times 80}}$$ = 0.35 g N2
(c) Molar mass of Ba(N3)2 = 221 g/mol
221 g of Ba(N3)2 produce 3 $$ \times $$ 28 g of N2
$$\therefore\,\,\,$$ 1 g of Ba(N3)2 produce = $${{3 \times 28} \over {221}}$$ = 0.38 g of N2
(d) Molar mass of NH3 = 17 g/mol
17 $$ \times $$ 2 NH3 produce 28 g of N2
$$\therefore\,\,\,$$ 1 g of NH3 produce = $${{28} \over {17 \times 2}}$$ = 0.823 g of N2
252g of (NH4)2 Cr2 O7 produce 28g mole of N2
$$\therefore\,\,\,$$ 1 g of (NH4)2 Cr2 O7 Produce = $${{28} \over {252}} = 0.111$$ g N2
(b) Molar mass of NH4 NO3 = 80 g/mol
2 $$ \times $$ 80 g of NH4 NO3 produce 28 $$ \times $$ 2g of N2
$$\therefore\,\,\,$$ 1 g of NH4 NO3 produce = $${{28 \times 2} \over {2 \times 80}}$$ = 0.35 g N2
(c) Molar mass of Ba(N3)2 = 221 g/mol
221 g of Ba(N3)2 produce 3 $$ \times $$ 28 g of N2
$$\therefore\,\,\,$$ 1 g of Ba(N3)2 produce = $${{3 \times 28} \over {221}}$$ = 0.38 g of N2
(d) Molar mass of NH3 = 17 g/mol
17 $$ \times $$ 2 NH3 produce 28 g of N2
$$\therefore\,\,\,$$ 1 g of NH3 produce = $${{28} \over {17 \times 2}}$$ = 0.823 g of N2
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