c. What is the amount of magnesium nitride (Mg3N2) produced if 0.225g of magnesium (Mg) is heated with 0.086 g of nitrogen gas (N2) in a closed container? The reaction obeyed the law of conservation of mass.


To determine the amount of magnesium nitride (\( \text{Mg}_3\text{N}_2 \)) produced, we need to examine the chemical reaction between magnesium (Mg) and nitrogen gas (\( \text{N}_2 \)).

### Step 1: Write the balanced chemical equation.

\[
3\text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2
\]

This tells us that 3 moles of magnesium react with 1 mole of nitrogen gas to produce 1 mole of magnesium nitride.

### Step 2: Calculate the moles of magnesium and nitrogen gas.

- **Moles of magnesium**:
  \[
  \text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} = \frac{0.225 \, \text{g}}{24.305 \, \text{g/mol}} \approx 0.00926 \, \text{mol}
  \]

- **Moles of nitrogen gas**:
  \[
  \text{Moles of N}_2 = \frac{\text{Mass of N}_2}{\text{Molar mass of N}_2} = \frac{0.086 \, \text{g}}{28.014 \, \text{g/mol}} \approx 0.00307 \, \text{mol}
  \]

### Step 3: Determine the limiting reactant.

From the balanced equation, 3 moles of magnesium react with 1 mole of nitrogen gas. Therefore, we need to compare the ratio of moles available for both reactants.

- Required ratio of Mg to \( \text{N}_2 \) is 3:1, so:
  \[
  \text{Required moles of Mg for } 0.00307 \, \text{mol of } \text{N}_2 = 3 \times 0.00307 = 0.00921 \, \text{mol of Mg}
  \]

Since we have 0.00926 moles of Mg, which is slightly more than the required 0.00921 moles, nitrogen gas (\( \text{N}_2 \)) is the **limiting reactant**.

### Step 4: Calculate the amount of \( \text{Mg}_3\text{N}_2 \) produced.

According to the balanced equation, 1 mole of \( \text{N}_2 \) produces 1 mole of \( \text{Mg}_3\text{N}_2 \). Therefore, the moles of \( \text{Mg}_3\text{N}_2 \) formed are equal to the moles of \( \text{N}_2 \).

\[
\text{Moles of Mg}_3\text{N}_2 = 0.00307 \, \text{mol}
\]

Now, calculate the mass of \( \text{Mg}_3\text{N}_2 \):

\[
\text{Mass of Mg}_3\text{N}_2 = \text{Moles of Mg}_3\text{N}_2 \times \text{Molar mass of Mg}_3\text{N}_2 = 0.00307 \, \text{mol} \times 100.928 \, \text{g/mol}
\]

\[
\text{Mass of Mg}_3\text{N}_2 \approx 0.3099 \, \text{g}
\]

### Final Answer:
The amount of magnesium nitride (\( \text{Mg}_3\text{N}_2 \)) produced is approximately **0.310 g**.

---

Would you like more details or have any questions? Here are five related questions you might find helpful:

1. How do we determine the limiting reactant in a chemical reaction?
2. What is the significance of the molar mass in stoichiometric calculations?
3. How does the law of conservation of mass apply to this reaction?
4. What are some common uses of magnesium nitride in industry?
5. How would the result change if the quantities of magnesium and nitrogen were reversed?

**Tip:** Always start by balancing the chemical equation before performing any stoichiometric calculations to ensure accuracy.

What is the amount of magnesium nitride (Mg3N2) produced if 0.225g of magnesium (Mg) is heated with 0.086 g of nitrogen gas (N2) in a closed container? The reaction obeyed the law of conservation of mass.

Learn how to calculate the mass of magnesium nitride (Mg3N2) produced from the reaction of 0.225g of magnesium and 0.086g of nitrogen gas. This solution explains the stoichiometric process, including the concepts of limiting reactants and molar mass, following the law of conservation of mass. Suitable for Grades 10-12.

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