Solved Using the law of conservation of mass, solve the | Chegg.com - Free Printable

Let's solve each of these problems using the Law of Conservation of Mass, which states:

> Mass of reactants = Mass of products

This means that in a chemical reaction, the total mass of the substances before the reaction (reactants) equals the total mass after the reaction (products), assuming no mass is lost to the surroundings.

We’ll go through each problem one by one.

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Problem 1:

Reaction:
N₂ + 3H₂ → 2NH₃

Given:
- 28 g of N₂
- 6 g of H₂

Find: Mass of NH₃ produced.

Solution:

According to the law of conservation of mass:

> Mass of reactants = Mass of products

So,
> Mass of NH₃ = Mass of N₂ + Mass of H₂
> = 28 g + 6 g = 34 g

✔ Answer: 34 g of NH₃

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Problem 2:

Reaction:
2H₂ + O₂ → 2H₂O

Given:
- 4 g of H₂
- 32 g of O₂

Find: Mass of H₂O produced.

Solution:

Again, apply conservation of mass:

> Mass of H₂O = Mass of H₂ + Mass of O₂
> = 4 g + 32 g = 36 g

✔ Answer: 36 g of H₂O

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Problem 3:

Reaction:
N₂ + O₂ → 2NO

Given:
- 28 g of N₂
- 32 g of O₂

Find: Mass of NO produced.

Solution:

> Mass of NO = Mass of N₂ + Mass of O₂
> = 28 g + 32 g = 60 g

✔ Answer: 60 g of NO

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Problem 4:

Reaction:
C + O₂ → CO₂

Given:
- 12 g of C
- 32 g of O₂

Find: Mass of CO₂ produced.

Solution:

> Mass of CO₂ = Mass of C + Mass of O₂
> = 12 g + 32 g = 44 g

✔ Answer: 44 g of CO₂

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✔ Summary of Answers:

1. 34 g of NH₃
2. 36 g of H₂O
3. 60 g of NO
4. 44 g of CO₂

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💡 Note: These calculations assume that the reactions go to completion and that the amounts given are stoichiometrically balanced (i.e., no limiting reactant issues). In real stoichiometry, we would check for limiting reactants, but since this exercise focuses on the law of conservation of mass, and the masses provided match the molar ratios (e.g., 28g N₂ = 1 mol, 6g H₂ = 3 mol, etc.), they are perfectly balanced.

Let me know if you'd like to see how to verify with mole calculations too!