Question

1. Ammonia gas may be formed by the reaction of nitrogen gas and hydrogen gas. When 10.0g of nitrogen is combined with 5.0g of hydrogen:
2. What is the limiting reagent?
3. What is the theoretical yield?
4. What is the mass of excess reagent remaining?

Answer 1

Ammonia is produced by reaction of nitrogen and and hydrogen.

N₂ (g) + 3H₂ (g) $\rightarrow$ 2NH₃ (g)

one mole of Nitrogen gas reacts with three moles Hydrogen gas.

for this reaction , N₂ (g) :  3H₂(g) = 1 : 3

We have, moles of Nitrogen gas (M.W. : 28 g ) = 10 g / 28 g/mol = 0.357 mole

  moles of Hydrogen gas (M.W. : 2 g ) = 5 g / 2 g/mol = 2.5 mole.

a. A limiting reagent is a substance in a chemical reaction, which is completely consumed during reaction, and its consumption stops reaction whatever amount of other reagents .

for this reaction required amounts    N₂ (g) :  3H₂(g) = 1 : 3

we have,   N₂ (g) :  3H₂(g) = 0.357 : 2.5 =  1 : 7

So, Nitrogen is limiting reagent.

b. Theoretical yield : It is yield obtained from complete conversion of limiting reagent .

We have 0.357 moles of limiting reagent Nitrogen , since 1 mol of N₂ (g) reacts with 3 moles of H₂(g) to produce 2 moles of Ammonia .

So, theoretical yield NH₃ = 2*0.357 moles = 0.714 moles ( M.W. NH₃ = 17 g /mol )

or  0.714 moles * 17 g /mol = 12.138 g.

c. Remaining Mass of excess reagent, H₂(g) :

Moles of H₂(g) reacted = 3*0.357 mole = 1.07 mole

Remaining = (2.5 - 1.07 ) moles = 1.43 moles

So, Remaining Mass of H₂(g) = 2.86 g.