Question

28 g of N2 and 6 g of H2 react to give 34 g of NH3. This statement illustrates the law of​

Answer 1

Answer:

The reaction for the formation of ammonia is N

2

+3H

2

⇔2NH

3

.

1 mole (28 g) of nitrogen will react with 3 moles (6g) of hydrogen to form 2 moles (34g) of ammonia if the reaction is 100% completed.

But only 17 g of ammonia are actually formed.

Hence, the reaction is 50% complete.

Hence, 0.5 mole (14 g) of nitrogen and 3 moles (3g) of hydrogen will remain unreacted at equilibrium. Hence, option C is correct.

Answer 2

28 g of $\[{N_2}\]$ and 6 g of $\[{H_2}\]$ react to give 34 g of $\[N{H_3}\]$. This statement illustrates the law of​ conservation of mass.

Explanation:

• According to the law of conservation of mass, the mass can neither be created nor destroyed in a chemical reaction.
• Therefore, the sum of the masses of the reactants must be equal to the sum of the masses of the products.
• According to the given statement, the reaction can be written as:

       $\[{N_2}(24\,g) + 3{H_2}(6\,g) \to 2N{H_3}(34\,g)\]$

• Since the mass of nitrogen is 28 g and the mass of hydrogen is 6 g, therefore, the sum of the masses of the reactants will be $\[28\,g + 6\,g = 34\,g\]$.
• The total mass of the product (ammonia) is also 34 g.
• Therefore, the total mass of the reactants is equal to the total mass of the product.
• Thus, the law of conservation of mass is satisfied.