Question

In the reaction N2 + 3H2 -----> 2 NH3 , determine the amount of NH3 formed if 2000g dinitrogen reacts with 1000g dihydrogen.​

Answer 1

$\huge{\green{\underline{\underline{Answer:-}}}}$

Given:

N2 + H2 ----> 2NH3

2000 gram N2 reacts with 1000 gram hydrogen

To find:

Amount of NH3 formed = ?

Solution:

N2 + 3H2 ----> 2NH3

Molar mass of N2 = 14+14 = 28 g

Moles = Mass/Molar mass

So, moles of nitrogen = 2000/28 = 71.42

Molar mass of H2 = 2g

Moles of Hydrogen = 1000/2 = 500.

Limiting reagent(LR) = Moles of reactant/ stoichiometric coefficient of reactant

So, For N2

LR = 71.42/1 = 71.42

For H2

LR = 500/3 = 166.6

So limiting reagent here is N2

Then

1 mole N2 gives 2 mole NH3

71.42 mole N2 gives 71.42 × 2/1 = 142.84 mole

Answer:

Therefore, 142.84 mole of NH3 is formed