Question

2g of nitrogen reacts with 4g of hydrogen to form ammonia . identify excess nd limiting reagent

Answer 1

Your answer is ----

The reaction is :-

N2 + H2 =====> NH3

firstly we balanced it , we get

N2 + 3H2 ======> 2NH3

Mol of N is :-

2/28 = 0.071 mol

Mol of H is :-

4/2 = 2Mol

according to equation

1 mol N2 react with 3mol H2 to give 2mol NH3

So,

0.071 mol N2 give 0.21 mol H2

But we have 2Mol H2

Therefore, H2 is the excess reagent and remain Means N2 is limiting reagent .

Answer 2

Mol. Wt. of N2- 28

Mol. Wt. of H2- 2

Since, we have 2g of N2 & 4g of H2

Thus,

No. of moles of N2- 2/28 = 0.07 mole

& No. of moles of H2- 4/2 = 2 moles

Now, from the stoichiometry of the reaction

N2 + 3H2 → 2NH3

We can see that, for 1 mole of N2 we require 3 moles of H2 to produce 2 moles of NH3 and here for 0.07 mole of N2 we have 2 moles of H2 i.e. we have 0.66 moles of excess H2 which will be unused. Thus for the given case N2 will be the limiting reagent.

u can calculate in this process also

we have 2g of N2 & 4g of H2

First you write the equation for formation of ammonia from hydrogen and nitrogen:

N2 + 3H2 → 2NH3

Note that nitrogen and hydrogen react in the ratio 1:3. The reactants that you have are in the ratio 1:2. Thus you have more nitrogen than you need to react with the hydrogen that you have - nitrogen is in excess (or, you don’t have enough hydrogen to react with all your nitrogen). You then write a balanced equation using the actual moles of the reagent that isn’t in excess (hydrogen):

1 *4/3N2 +3 * 4/3 H2 → 2 *4/3 NH3

4/3N2 + 4 H2 → 8/3 NH3

4/3N2 =1.33 moles needed

we have 2 mole

Thank you.

if it is helpful to u  plz mark brainly