Question

28 g of nitrogen is mixed with 12 g of hydrogen to form ammonia as per the reaction, n2 + 3 h2 → 2nh3. which is the limiting reagent in this reaction?

Answer 1

Number of moles = Weight (in grams)/Molecular Wt.

Mol. Wt. of N2- 28

Mol. Wt. of H2- 2

Since, we have 28g of N2 & 12g of H2

Thus,

No. of moles of N2- 28/28 = 1 mole

& No. of moles of H2- 12/2 = 6 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 1 mole of N2 we have 6 moles of H2 i.e. we have 3 moles of excess H2 which will be unused. Thus for the given case N2 will be the limiting reagent.

Answer 2

Answer:

The limiting reagent is Nitrogen(N2)

Explanation:

Limiting reagent in a reaction is one which is totally consumed before any other reactant.

To find limiting reagent we first need to find no.of moles of reactants and products

$no.of \: moles = \frac{weight}{molar \: mass}$

Given that,

Weight of Nitrogen=28gm

$no.of \: moles = \frac{28}{28} = 1 \: mol$

Weight of Hydrogen=12gm

$no.of \: moles = \frac{12}{2} = 6mol$

Now we write the equation of the chemical reaction

N2 + 3H2 → 2NH3

From this reaction,we observe that to produce 2 moles of Ammonia, 1 mole of Nitrogen and 3 moles of Hydrogen are required

In the given question,instead of 3 moles,we have 6 moles of Hydrogen,therefore 3 moles of Hydrogen will be in excess,hence it can't be a limiting reagent

Therefore the limiting reagent is Nitrogen(N2)

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