Question

Equal amount of of N_{​​​​​​2} and H_{​​​​​​2} react to form ammonia under favourable conditions then the limiting reagent is????

Answer 1

Answer:

Convert mass quantities to moles and divide by respective coefficients of balanced equation. The smaller number is the limiting reagent.

Explanation:

3H2+N2→2NH3

Example: Assume Given 100 g of each reactant, what's the limiting reagent?

moles of H2=(100g/2gmol−1)=50 moles & Divide by '3' => 16.7

moles of N2=(100g/28gmol−1)=3.57 moles & Divide by '1' => 3.57

The limiting reagent is N2 because after dividing by respective coefficients => 3.57 is smaller than 16.7 and H2 will be in excess; after all N2 is consumed in the reaction.