Question

60. Calculate the mass of NH3, produced when 14 g of nitrogen is burnt in
24 g of hydrogen​

Answer 1

$\boxed{\bold{\sf{\Large{\red{N_2+3H_2=2NH_3}}}}}$

$\rule{200}{2}$

Mass of Nitrogen given = 14g

Molar mass of Nitrogen = 28 g

Number of moles =14/28 =0.5 moles

Mass of Hydrogen given=24g

Molar mass of Hydrogen=2 g

Number of Moles =24/2 = 12 moles

$\rule{200}{2}$

$Mole\:Ratio=\dfrac{moles}{stoichiometric\:\:coefficient}$

Thus,

Mole ratio of Nitrogen = 0.5

Mole ratio of Hydrogen = 4

Thus,

Nitrogen is Limiting Reagent.

$\rule{200}{2}$

1 mole of nitrogen molecule produces 2 moles of Ammonia

(From the Reaction at the top)

Therefore:-

0.5 moles of Nitrogen will produce 1 mole of Ammonia.

$\rule{200}{2}$

Molecular mass of Ammonia = 14+3 = 17 u

Molar mass of Ammonia = 17 Grams

Thus,

Mass of Ammonia Produced will be equal to :-

$=1*17\\ \\ \\=17\:Grams$

$\rule{200}{2}$