Question

How many grams of NH3 can be prepared from 77.3 grams of N2 and 14.2 grams of H2?
(Hint: N, +3H, → 2NH3)
80.4 g
79.7 g
47.0 g
120.0 g​

Answer 1

Answer:

the answer is

80.4g

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Answer 2

Given:

N₂ + 3H₂ → 2NH₃

The mass of N₂ = 77.3 g

The mass of H₂ = 14.2 g

To Find:

The amount of NH₃ evolved.

Solution

• As per the equation given:

                                 N₂ + 3H₂ → 2NH₃

1 mole of N₂ reacts with 3 moles of H₂ to evolve 2 moles of NH₃.

• For finding the answer to this question, we will use the following steps:

1. Find the limiting reagent.

What is the limiting reagent?

"The limiting reagent in a chemical reaction is a reactant that is consumed when the chemical reaction is completed."

• This means that the limiting reagent will decide the amount of product formed because the reaction will not go any further without it.

Now, the molar mass of N₂ = 28 g

and the molar mass of H₂ = 2 g

• So, the number of moles of N₂ = $\frac{77.3}{28}$ = 2.76 moles
• The number of moles of H₂ = $\frac{14.2}{2}$ = 7.1 moles

• Now, we know that 1 mole of N₂ reacts with 3 moles of H₂,

so 2.76 moles of N₂ will react with:

                  ⇒(2.76 × 3) moles of H₂

                  ⇒ 8.28 moles of H₂

Since the required number of moles of H₂ < available moles,

∴ H₂ is the limiting reagent.

2. Calculate the number of moles of the product formed as per the limiting reagent.

• Since we only have 7.1 moles of H₂,

                                 $\frac{1}{3}$× 7.1

                              ⇒ 2.36  moles of N₂ will react.

• We have:

                             N₂    +     3H₂    →     2NH₃

Initial moles:       2.76           7.1              0

Final moles:         0.4             0               x

• To find x, we will apply the unitary method:

3 moles of H₂ evolves 2 moles of NH₃

Hence, 7.1 moles of H₂ will evolve:

                             $\frac{2}{3}$ × 7.1

                        ⇒ ≈4.73 moles

3. Find the mass of the product formed.

• The molar mass of NH₃ = 17g

Hence, mass of 4.73 moles = 17 × 4.73 = 80.4 g

Thus, 80.4 g of NH₃ can be prepared from 77.3 grams of N₂ and 14.2 grams of H₂.

So, option 1 is correct.