Verified Solution • 2025–26 Session moderate

QJEEM18Chc960: JEE Main 2018 Chemistry - Class 11 JEEM18Chc960 Solution: For per gram of reactant, the maximum quantity of N 2 gas is…

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)

Multiple Choice Options

  • A (NH4)2Cr2O7(s) →\to N2(g) + 4H2O(g) + Cr2O3(s)
  • B 2NH4NO3(s) →\to 2 N2(g) + 4H2O(g) + O2(g)
  • C Ba(N3)2(s) →\to Ba(s) + 3N2(g)
  • D 2NH3(g) →\to N2(g) + 3H2(g)

Step-by-Step Verified Solution

(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 = 28252=0.111{{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 = 28×22×80{{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 = 3×28221{{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 = 2817×2{{28} \over {17 \times 2}} = 0.823 g of N2

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