Question

50.0kg of N2(g) and 10.0kg of H2(g) are mixed to produce NH3(g) calculate the moles of NH3(g) formed. Identify the limiting reagent in the production of NH3 in this solution

Answer 1

Answer:

Amount of NH3 formed

= approx 56,200 mol

Limiting reagent = H2

Explanation:

From balance chemical equation:.

N2 + 3H2 -------> 2NH3

It is obvious that 1mol of N2 react with 3mol of H2 to form 2mol of NH3.

It means that 28 gram of N2 require 6 gram of H2 to form 34 gram of NH3

THUS 10,000 gram of H2 require 46,284 gram of N2 to form 56,202 of NH3.

Also as the amoint of N2 is more than the required value so this is in excess, while the Hydrogen (H2) is limiting reagent of the reaction.