Question

Consider the following reaction: 2 NO(g) + 5 H2(g) ? 2 NH3(g) + 2 H2O(g) Which set of solution maps would be needed to calculate the maximum amount of ammonia (NH3), in grams, that can be synthesized from 45.8 g of nitrogen monoxide (NO) and 12.4 g of hydrogen (H2)? I. g NO ? mol NO ? mol NH3 ? g NH3 II. g H2 ? mol H2 ? mol NH3 ? g NH3 III. g NO ? mol NO ? mol H2O ? g H2O IV. g H2 ? mol H2 ? mol H2O ? g H2O. solutions maps I and II solutions maps I and V solutions maps I and III solutions maps II and III

Answer 1

Ans :- I. g NO → mol NO → mol NH₃ → g NH₃is the correct answer.

Explanation :-

Given moles of NO = mass / molar mass = 45.8g / 30 g/mol = 1.527 mol

Given moles of H₂ = 12.4 g / 2g/mol = 6.2 mol

Given reaction is :

2 NO(g) + 5 H₂(g) -------------> 2 NH₃ (g) + 2 H₂O(g)

From the balance chemical equation, it is cleared that,

5 mol of H₂ reacts with = 2 mol of NO

So,

6.2 mol (given) of H₂ reacts with = 2 mol x 6.2 mol / 5 mol of NO = 2.48 mol of NO

Since, the given moles of NO i.e. 1.527 mol are less, therefore NO is the limiting reagent.

Now,

2 mol of limiting reagent i.e. NO gives = 2 mol of NH₃

So,

1.527 mol (given) of NO gives = 1.527 mol of NH₃

and

Mass of NH₃ formed = moles x gram molar mass = 1.527 mol x 17 g/mol = 25.99 g