Question

56 g of Nitrogen reacts with 22 g of hydrogen to form ammonia.Find

I) The limiting reagent of the reaction.

2) The amount of reactant in excess

3) The amount of ammonia formed.

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Answer 1

Answer :-

1) Limiting reagent is Nitrogen .

2) Hydrogen is present in excess .

3) 68 grams of ammonia is formed .

Explanation :-

The balanced chemical equation for the reaction is :-

N₂ + 3H₂ → 2NH₃

Moles of N₂ :-

= Given Mass/Molar mass

= 56/28

= 2 moles

Moles of H₂ :-

= Given Mass/Molar mass

= 22/2

= 11 moles

________________________________

From the balanced equation, we can understand that 1 mole of N₂ will combine with 3 moles of H₂ . Then, 2 moles of N₂ will combine with 6 moles of H₂ .

So, nitrogen (N₂) is the limiting reagent and it will control the amount of product .

Hydrogen (H₂) is present in excess .

________________________________

Now, again from the balanced equation :-

∵ 28 g of N₂ → 34 g of NH₃

∴ 56 g of N₂ will give :-

= 56 × 34/28

= 2 × 34

= 68 g of NH₃.

Answer 2

Answer:

Given :-

• 56 g of Nitrogen reacts with 22 g of hydrogen to form ammonia.

To Find :-

• The limiting reagent of the reaction.
• The amount of reactant in excess.
• The amount of ammonia formed.

Solution :-

$\clubsuit$ Balanced Equation :

$\bigstar\: \: \sf\boxed{\bold{N_2 + 3H_2 \longrightarrow 2NH_3}}$

Now, we have to find the number of moles of nitrogen and hydrogen :

${\normalsize{\bold{\purple{\underline{\leadsto\: In\: case\: of\: Nitrogen\: :-}}}}}$

$\implies \sf \bold{\pink{Number\: of\: moles =\: \dfrac{Mass}{Molar\: Mass}}}$

$\implies \sf Number\: of\: moles =\: \dfrac{\cancel{56}}{\cancel{28}}$

$\implies \sf Number\: of\: moles =\: \dfrac{2}{1}$

$\implies \sf\bold{\green{Number\: of\: moles =\: 2\: moles}}$

${\normalsize{\bold{\purple{\underline{\leadsto \: In\: case\: of\: hydrogen\: :-}}}}}$

$\implies \sf Number \: of\: moles =\: \dfrac{\cancel{22}}{\cancel{2}}$

$\implies \sf Number\: of\: moles =\: \dfrac{11}{1}$

$\implies \sf \bold{\green{Number\: of\: moles =\: 11\: moles}}$

Again,

$\clubsuit$ Balanced Equation :

$\bigstar\: \: \: \sf\boxed{\bold{28\: g\: of\: N_2 \longrightarrow 34\: g\: of\: NH_3}}$

Given that :

$\mapsto$ 56 g of Nitrogen

$\implies \sf 56 \times \dfrac{34}{28}$

$\implies \sf \dfrac{1904}{28}$

$\implies \sf \dfrac{68}{1}$

$\implies \sf\bold{\red{68\: g}}$

$\rule{150}{3}$

1) The limiting reagent of the reaction :

➲ The limiting reagent of the reaction is Nitrogen

2) The amount of reactant in excess :

➲ The amount of reactant in excess is Hydrogen

3) The amount of ammonia formed :

➲ The amount of ammonia formed is 68 grams