Question

28g of N2 and 6g of H2 were mixed.At equilibrium 17g of NH3 was produced.The weight of N2 and H2 produced at equilibrium are respectively
1)11g,0g 2)1g ,3g 3)14g,3g. 5)11g,3g

Answer 1

28g of N2 and 6g of H2 were mixed.

molar mass of N2 = 28 g/mol

so, number of mole of N2 is mixed = 28/28 = 1

molar mass of H2 = 2 g/mol

so, number of mole of H2 is mixed = 6/2 = 3

see reaction,

N2 + 3H2 => 2NH3

at equilibrium ,

number of mole of N2 = 1 - x

number of mole of H2 = 3 - 3x

number of mole of NH3 = 2x

so, mass of NH3 is produced = mole × molar mass of NH3

= 2x × (14 +3) = 34x

given, mass = 17g = 34x

so, x = 0.5

so, at equilibrium,

number of mole of N2 = 1 - 0.5 = 0.5

hence,mass of N2 = 0.5 × 28 = 14g

number of mole of H2 = 3 - 3 × 0.5

= 1.5

hence, mass of H2 = 1.5 × 2 = 3g

hence, answer is 14g, 3g so, option (3) is correct choice.

Answer 2

Answer:

(3) 14 g, 3 g

here's the answer and hope it helps you.