If 4 moles of nitrogen reacts with 20 moles of hydrogen how many nh3 m

1.	If 4 moles of nitrogen reacts with 20 moles of hydrogen how many nh3 moles are produced
Answer» Answer: GIVEN: Reaction of Hydrogen and Nitrogen is given to form Ammonia. 4 moles of hydrogen reacts with 20 moles of hydrogen. To find: Moles of ammonia PRODUCED. Concept: A reaction forming products is governed according to the amount of limiting REAGENT PRESENT in the system. Calculation: $\boxed{ \red{N_{2} + 3H_{2} \: \rightarrow \: 2NH_{3}}}$ So we can say that in ideal conditions; 1 mole of Nitrogen react with 3 mole of hydrogen to form 2 moles of Ammonia. So, here nitrogen is the limiting reagent and hydrogen is the excess reagent. Because as per Unitary method, 4 moles of nitrogen shall react with 12 moles of hydrogen (but 20 moles have been provided). So moles of Ammonia produced = 4 × 2 = 8 moles. Amount (in gm) of ammonia produced = 8 × 17 = 136 gm. So final answer : $\boxed{ \blue{amount \: = 136 \: gm}}$

If 4 moles of nitrogen reacts with 20 moles of hydrogen how many nh3 moles are produced

Answer»

Reaction of Hydrogen and Nitrogen is given to form Ammonia.

4 moles of hydrogen reacts with 20 moles of hydrogen.

Moles of ammonia PRODUCED.

A reaction forming products is governed according to the amount of limiting REAGENT PRESENT in the system.

$\boxed{ \red{N_{2} + 3H_{2} \: \rightarrow \: 2NH_{3}}}$

So we can say that in ideal conditions; 1 mole of Nitrogen react with 3 mole of hydrogen to form 2 moles of Ammonia.

So, here nitrogen is the limiting reagent and hydrogen is the excess reagent.

Because as per Unitary method,

4 moles of nitrogen shall react with 12 moles of hydrogen (but 20 moles have been provided).

So moles of Ammonia produced

= 4 × 2

= 8 moles.

Amount (in gm) of ammonia produced

= 8 × 17

= 136 gm.

So final answer :

$\boxed{ \blue{amount \: = 136 \: gm}}$