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JEE Mains Previous Paper 1 (Held On: 09 Apr 2019 Shift 1)

Option 1 : 56 g of N_{2} + 10 g of H_{2}

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

Stoichiometry is an important concept in chemistry that helps us use balanced chemical equations to calculate amounts of reactants and products. Here, we make use of ratios from the balanced equation. In general, all the reactions that take place are dependent on one main factor, how much substance is present.

**Calculation:**

According to the stoichiometry of balanced equation

a) 56 g of N_{2} + 10 g of H_{2}

= 2 mole of N_{2 }+ 5 mole of H_{2}

= 2/1 mole of N_{2 }+ 5/3 mole of H_{2}

= 2 mole of N_{2 }+ 1.66 mole of H_{2}

b) 35 g of N_{2} + 8 g of H_{2}

= 35/28 mole of N_{2 }+ 4 mole of H_{2}

= 35/28 mole of N_{2 }+ 4 mole of H_{2}

= 35/28 mole of N_{2 }+ 4/3 mole of H_{2}

= 1.25 mole of N_{2 }+ 1.33 mole of H_{2}

c) 28 g of N_{2} + 6 g of H_{2}

= 1 mole of N_{2 }+ 3 mole of H_{2}

= 1/1 mole of N_{2 }+ 3/3 mole of H_{2}

= 1 mole of N_{2 }+ 1 mole of H_{2}

d) 14 g of N_{2} + 4 g of H_{2}

= 1/2 mole of N_{2 }+ 2 mole of H_{2}

= 1/2 mole of N_{2 }+ 2/3 mole of H_{2}

= 1/2 mole of N_{2 }+ 0.66 mole of H_{2}

28 g N_{2} react with 6 g H_{2}

\(\mathop {{N_2}}\limits_{1\mathop {mole}\limits_{28g} \;} + \mathop {3{H_2}}\limits_{3\mathop {mole}\limits_{6g} \;} \to 2N{H_3}\;\)

∴ For 56 g of N_{2}, 12 g of H_{2} is required.

N_{2} (g) + 3H_{2} (g) → 2NH_{3} (g)