Question

Determine
the
quantity
of
in
STARTING AMOUNT





18.0
g
HgBr₂
×
1
mol
HgBr₂
360.40
g
HgBr₂
×
2
mol
Br
1
mol
HgBr₂
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Answer 1

To determine the quantity of bromine (\( \text{Br} \)) produced from the starting amount of \( \text{HgBr}_2 \), you can use stoichiometry.

1. **Identify the molar mass of \( \text{HgBr}_2 \)**:
- The molar mass of mercury (Hg) is approximately 200.59 g/mol.
- The molar mass of bromine (Br) is approximately 79.90 g/mol.

Therefore, the molar mass of \( \text{HgBr}_2 \) is:
\[
200.59 \, \text{g/mol} + (2 \times 79.90 \, \text{g/mol}) = 200.59 \, \text{g/mol} + 159.80 \, \text{g/mol} = 360.39 \, \text{g/mol}
\]
(We will use \( 360.40 \, \text{g/mol} \) as provided.)

2. **Convert grams of \( \text{HgBr}_2 \) to moles**:
\[
\text{moles of } \text{HgBr}_2 = \frac{18.0 \, \text{g}}{360.40 \, \text{g/mol}} \approx 0.0499 \, \text{mol}
\]

3. **Use the stoichiometry of the reaction**:
The decomposition of \( \text{HgBr}_2 \) usually follows the equation:
\[
\text{HgBr}_2 \rightarrow \text{Hg} + \text{Br}_2
\]
From the balanced reaction, 1 mole of \( \text{HgBr}_2 \) produces 2 moles of \( \text{Br} \).

Therefore, the moles of \( \text{Br} \) produced will be:
\[
\text{moles of } \text{Br} = 0.0499 \, \text{mol} \times 2 = 0.0998 \, \text{mol}
\]

4. **Convert moles of bromine to grams**:
\[
\text{mass of } \text{Br} = 0.0998 \, \text{mol} \times 159.80 \, \text{g/mol} \approx 15.95 \, \text{g}
\]

### Final Answer:
- The quantity of bromine produced from 18.0 g of \( \text{HgBr}_2 \) is approximately **15.95 g**.