Question

Consider the reaction, N2(g) + 3 H2(g) → 2 NH3(g). Suppose 56 g of N2(g) reacts with 18 g of H2(g).

d. What is the limiting reactant?

e. When the reaction is complete, how many grams of NH3(g) are produced?

f. When the reaction is complete, how many grams of N2(g) remain?

g. When the reaction is complete, how many grams of H2(g) remain?

h. If the reaction actually produced 55 g of NH3(g), what is the percent yield?

Answer 1

d. Limiting Reagent is the reagent that is present in less quantity in a reaction and get fully consumed in a chemical reaction. Due to which, product formation gets limited.

Here in the reaction, 1 mole of N₂ reacts with 3 moles of H₂ means we need 14gm of N₂ needed to react with 2*3=6 gm of H₂.

We have 56 gm of N₂ means 2 moles (mol. mass=28g/mol) and 18 gm H₂ means 9 moles(mol. mass=2g/mol)

1 mole of N₂ reacts with 3 moles of H₂

2 moles will react with 6 moles of H₂.

but we have 9 moles of H₂ means we have an excess quantity of H₂, so limiting reagent is N₂ as its quantity is limited.

e. According to reaction, 1 mole of N₂ and 3 moles of H₂ produce 2 moles of NH₃ means 34gm(mol. mass= 17g/mol)

we have 2 moles of N₂, so it will produce 4 moles of NH₃ means 4*17= 68 gm NH₃.

so, 68 gm NH₃ will be produced.

f. As N₂ is limiting reagent, it will get fully consumed in the reaction.

g. when the reaction is completed, 6 moles(12g) of H₂ are consumed so we are left with 6 gm of H₂.

h. according to reaction, 68 gm NH₃ will produce but according to question 55g of NH₃ is produced.

Percent Yield= (Actual yield / Theoretical Yield) * 100

= (55/68)*100 = 80.8%

so, percent yield is 80.8%.