Solving chemistry problems using the law of conservation of mass.

To solve these problems, we will use the law of conservation of mass, which states that the total mass of reactants must equal the total mass of products in a chemical reaction. Additionally, we will use stoichiometry to determine the masses of the products formed based on the given masses of the reactants.

Problem 1:

Reaction: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)

Given:
- 28 g of \( \text{N}_2 \)
- 6 g of \( \text{H}_2 \)

Task: Determine how many grams of \( \text{NH}_3 \) are produced.

#### Step 1: Calculate the molar masses.
- Molar mass of \( \text{N}_2 \): \( 2 \times 14.01 = 28.02 \, \text{g/mol} \)
- Molar mass of \( \text{H}_2 \): \( 2 \times 1.01 = 2.02 \, \text{g/mol} \)
- Molar mass of \( \text{NH}_3 \): \( 14.01 + 3 \times 1.01 = 17.03 \, \text{g/mol} \)

#### Step 2: Determine the moles of reactants.
- Moles of \( \text{N}_2 \):
\[
\text{Moles of } \text{N}_2 = \frac{28 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1 \, \text{mol}
\]
- Moles of \( \text{H}_2 \):
\[
\text{Moles of } \text{H}_2 = \frac{6 \, \text{g}}{2.02 \, \text{g/mol}} \approx 2.97 \, \text{mol}
\]

#### Step 3: Use the balanced equation to find the limiting reactant.
The balanced equation shows:
\[
1 \, \text{mol of } \text{N}_2 \text{ reacts with } 3 \, \text{mol of } \text{H}_2
\]
- Required moles of \( \text{H}_2 \) for 1 mol of \( \text{N}_2 \): 3 mol
- Available moles of \( \text{H}_2 \): 2.97 mol

Since 2.97 mol of \( \text{H}_2 \) is less than the required 3 mol, \( \text{H}_2 \) is the limiting reactant.

#### Step 4: Calculate the moles of \( \text{NH}_3 \) produced.
From the balanced equation:
\[
3 \, \text{mol of } \text{H}_2 \text{ produces } 2 \, \text{mol of } \text{NH}_3
\]
- Moles of \( \text{NH}_3 \) produced by 2.97 mol of \( \text{H}_2 \):
\[
\text{Moles of } \text{NH}_3 = \frac{2}{3} \times 2.97 \approx 1.98 \, \text{mol}
\]

#### Step 5: Calculate the mass of \( \text{NH}_3 \).
- Mass of \( \text{NH}_3 \):
\[
\text{Mass of } \text{NH}_3 = 1.98 \, \text{mol} \times 17.03 \, \text{g/mol} \approx 33.73 \, \text{g}
\]

Answer for Problem 1:
\[
\boxed{34 \, \text{g}}
\]

---

Problem 2:

Reaction: \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

Given:
- 4 g of \( \text{H}_2 \)
- 32 g of \( \text{O}_2 \)

Task: Determine how many grams of \( \text{H}_2\text{O} \) are produced.

#### Step 1: Calculate the molar masses.
- Molar mass of \( \text{H}_2 \): \( 2.02 \, \text{g/mol} \)
- Molar mass of \( \text{O}_2 \): \( 2 \times 16.00 = 32.00 \, \text{g/mol} \)
- Molar mass of \( \text{H}_2\text{O} \): \( 2 \times 1.01 + 16.00 = 18.02 \, \text{g/mol} \)

#### Step 2: Determine the moles of reactants.
- Moles of \( \text{H}_2 \):
\[
\text{Moles of } \text{H}_2 = \frac{4 \, \text{g}}{2.02 \, \text{g/mol}} \approx 1.98 \, \text{mol}
\]
- Moles of \( \text{O}_2 \):
\[
\text{Moles of } \text{O}_2 = \frac{32 \, \text{g}}{32.00 \, \text{g/mol}} = 1 \, \text{mol}
\]

#### Step 3: Use the balanced equation to find the limiting reactant.
The balanced equation shows:
\[
2 \, \text{mol of } \text{H}_2 \text{ reacts with } 1 \, \text{mol of } \text{O}_2
\]
- Required moles of \( \text{H}_2 \) for 1 mol of \( \text{O}_2 \): 2 mol
- Available moles of \( \text{H}_2 \): 1.98 mol

Since 1.98 mol of \( \text{H}_2 \) is less than the required 2 mol, \( \text{H}_2 \) is the limiting reactant.

#### Step 4: Calculate the moles of \( \text{H}_2\text{O} \) produced.
From the balanced equation:
\[
2 \, \text{mol of } \text{H}_2 \text{ produces } 2 \, \text{mol of } \text{H}_2\text{O}
\]
- Moles of \( \text{H}_2\text{O} \) produced by 1.98 mol of \( \text{H}_2 \):
\[
\text{Moles of } \text{H}_2\text{O} = 1.98 \, \text{mol}
\]

#### Step 5: Calculate the mass of \( \text{H}_2\text{O} \).
- Mass of \( \text{H}_2\text{O} \):
\[
\text{Mass of } \text{H}_2\text{O} = 1.98 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 35.70 \, \text{g}
\]

Answer for Problem 2:
\[
\boxed{36 \, \text{g}}
\]

---

Problem 3:

Reaction: \( \text{N}_2 + \text{O}_2 \rightarrow 2\text{NO} \)

Given:
- 28 g of \( \text{N}_2 \)
- 32 g of \( \text{O}_2 \)

Task: Determine how many grams of \( \text{NO} \) are produced.

#### Step 1: Calculate the molar masses.
- Molar mass of \( \text{N}_2 \): \( 28.02 \, \text{g/mol} \)
- Molar mass of \( \text{O}_2 \): \( 32.00 \, \text{g/mol} \)
- Molar mass of \( \text{NO} \): \( 14.01 + 16.00 = 30.01 \, \text{g/mol} \)

#### Step 2: Determine the moles of reactants.
- Moles of \( \text{N}_2 \):
\[
\text{Moles of } \text{N}_2 = \frac{28 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1 \, \text{mol}
\]
- Moles of \( \text{O}_2 \):
\[
\text{Moles of } \text{O}_2 = \frac{32 \, \text{g}}{32.00 \, \text{g/mol}} = 1 \, \text{mol}
\]

#### Step 3: Use the balanced equation to find the limiting reactant.
The balanced equation shows:
\[
1 \, \text{mol of } \text{N}_2 \text{ reacts with } 1 \, \text{mol of } \text{O}_2
\]
- Both reactants are present in equal amounts (1 mol each), so neither is limiting.

#### Step 4: Calculate the moles of \( \text{NO} \) produced.
From the balanced equation:
\[
1 \, \text{mol of } \text{N}_2 \text{ and } 1 \, \text{mol of } \text{O}_2 \text{ produce } 2 \, \text{mol of } \text{NO}
\]
- Moles of \( \text{NO} \) produced:
\[
\text{Moles of } \text{NO} = 2 \, \text{mol}
\]

#### Step 5: Calculate the mass of \( \text{NO} \).
- Mass of \( \text{NO} \):
\[
\text{Mass of } \text{NO} = 2 \, \text{mol} \times 30.01 \, \text{g/mol} = 60.02 \, \text{g}
\]

Answer for Problem 3:
\[
\boxed{60 \, \text{g}}
\]

---

Problem 4:

Reaction: \( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)

Given:
- 12 g of \( \text{C} \)
- 32 g of \( \text{O}_2 \)

Task: Determine how many grams of \( \text{CO}_2 \) are produced.

#### Step 1: Calculate the molar masses.
- Molar mass of \( \text{C} \): \( 12.01 \, \text{g/mol} \)
- Molar mass of \( \text{O}_2 \): \( 32.00 \, \text{g/mol} \)
- Molar mass of \( \text{CO}_2 \): \( 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \)

#### Step 2: Determine the moles of reactants.
- Moles of \( \text{C} \):
\[
\text{Moles of } \text{C} = \frac{12 \, \text{g}}{12.01 \, \text{g/mol}} \approx 1 \, \text{mol}
\]
- Moles of \( \text{O}_2 \):
\[
\text{Moles of } \text{O}_2 = \frac{32 \, \text{g}}{32.00 \, \text{g/mol}} = 1 \, \text{mol}
\]

#### Step 3: Use the balanced equation to find the limiting reactant.
The balanced equation shows:
\[
1 \, \text{mol of } \text{C} \text{ reacts with } 1 \, \text{mol of } \text{O}_2
\]
- Both reactants are present in equal amounts (1 mol each), so neither is limiting.

#### Step 4: Calculate the moles of \( \text{CO}_2 \) produced.
From the balanced equation:
\[
1 \, \text{mol of } \text{C} \text{ and } 1 \, \text{mol of } \text{O}_2 \text{ produce } 1 \, \text{mol of } \text{CO}_2
\]
- Moles of \( \text{CO}_2 \) produced:
\[
\text{Moles of } \text{CO}_2 = 1 \, \text{mol}
\]

#### Step 5: Calculate the mass of \( \text{CO}_2 \).
- Mass of \( \text{CO}_2 \):
\[
\text{Mass of } \text{CO}_2 = 1 \, \text{mol} \times 44.01 \, \text{g/mol} = 44.01 \, \text{g}
\]

Answer for Problem 4:
\[
\boxed{44 \, \text{g}}
\]

---

Final Answers:

1. \( \boxed{34 \, \text{g}} \)
2. \( \boxed{36 \, \text{g}} \)
3. \( \boxed{60 \, \text{g}} \)
4. \( \boxed{44 \, \text{g}} \)