Determine the number of grams of silver bromide formed when 3.50 grams of aluminum bromide reacts with 4.00 grams of silver nitrate.

To determine the number of grams of silver bromide (\( \text{AgBr} \)) formed when aluminum bromide (\( \text{AlBr}_3 \)) reacts with silver nitrate (\( \text{AgNO}_3 \)), we need to follow a series of steps using stoichiometry.

The balanced chemical equation for the reaction is:

\[
\text{AlBr}_3 + 3 \text{AgNO}_3 \rightarrow 3 \text{AgBr} + \text{Al(NO}_3)_3
\]

### Step 1: Calculate the molar masses
- Molar mass of \( \text{AlBr}_3 \):  
  \[
  \text{Al} = 26.98 \, \text{g/mol}, \quad \text{Br} = 79.90 \, \text{g/mol}
  \]
  \[
  \text{Molar mass of AlBr}_3 = 26.98 + 3 \times 79.90 = 266.68 \, \text{g/mol}
  \]

- Molar mass of \( \text{AgNO}_3 \):  
  \[
  \text{Ag} = 107.87 \, \text{g/mol}, \quad \text{N} = 14.01 \, \text{g/mol}, \quad \text{O} = 16.00 \, \text{g/mol}
  \]
  \[
  \text{Molar mass of AgNO}_3 = 107.87 + 14.01 + 3 \times 16.00 = 169.88 \, \text{g/mol}
  \]

- Molar mass of \( \text{AgBr} \):  
  \[
  \text{Ag} = 107.87 \, \text{g/mol}, \quad \text{Br} = 79.90 \, \text{g/mol}
  \]
  \[
  \text{Molar mass of AgBr} = 107.87 + 79.90 = 187.77 \, \text{g/mol}
  \]

### Step 2: Calculate moles of reactants
- Moles of \( \text{AlBr}_3 \):
  \[
  \text{Moles of AlBr}_3 = \frac{3.50 \, \text{g}}{266.68 \, \text{g/mol}} = 0.01313 \, \text{mol}
  \]

- Moles of \( \text{AgNO}_3 \):
  \[
  \text{Moles of AgNO}_3 = \frac{4.00 \, \text{g}}{169.88 \, \text{g/mol}} = 0.02355 \, \text{mol}
  \]

### Step 3: Determine the limiting reactant
The reaction requires 3 moles of \( \text{AgNO}_3 \) for every 1 mole of \( \text{AlBr}_3 \), so we calculate the required amount of \( \text{AgNO}_3 \) for 0.01313 mol of \( \text{AlBr}_3 \):

\[
\text{Required moles of AgNO}_3 = 3 \times 0.01313 = 0.03939 \, \text{mol}
\]

Since we only have 0.02355 mol of \( \text{AgNO}_3 \), it is the limiting reactant.

### Step 4: Calculate moles of \( \text{AgBr} \) formed
According to the balanced equation, 3 moles of \( \text{AgNO}_3 \) produce 3 moles of \( \text{AgBr} \), so the moles of \( \text{AgBr} \) formed are equal to the moles of \( \text{AgNO}_3 \) available:

\[
\text{Moles of AgBr} = 0.02355 \, \text{mol}
\]

### Step 5: Calculate the mass of \( \text{AgBr} \) formed
\[
\text{Mass of AgBr} = 0.02355 \, \text{mol} \times 187.77 \, \text{g/mol} = 4.42 \, \text{g}
\]

### Conclusion:
The mass of silver bromide (\( \text{AgBr} \)) formed is **4.42 grams**.

Would you like further details on any step?

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Here are five questions related to this problem:
1. What would happen if more silver nitrate were available than aluminum bromide?
2. How would you identify the limiting reactant if both reactants were in excess?
3. Why is it important to balance the chemical equation before performing stoichiometry?
4. Can this reaction yield any byproducts other than aluminum nitrate?
5. How would changing the initial mass of aluminum bromide affect the outcome?

**Tip**: When working with chemical reactions, always ensure that the equation is balanced before performing any calculations!

Learn how to calculate the grams of silver bromide formed when 3.50 grams of aluminum bromide reacts with 4.00 grams of silver nitrate using stoichiometric principles. This solution explains molar mass, limiting reactant concepts, and detailed calculations. Suitable for high school chemistry students.

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