To determine how many moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]) are formed when 183.5 grams of barium hydroxide ([tex]\( \text{Ba(OH)}_2 \)[/tex]) react with excess hydrobromic acid ([tex]\( \text{HBr} \)[/tex]), we follow these steps:
1. Calculate the molar mass of barium hydroxide ([tex]\( \text{Ba(OH)}_2 \)[/tex]):
- Barium (Ba) has an atomic mass of approximately 137.34 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
- Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.
Therefore, the molar mass of [tex]\( \text{Ba(OH)}_2 \)[/tex] is:
[tex]\[
\text{Molar mass of Ba(OH)}_2 = 137.34 + 2(\text{1 oxygen} \times 16.00 + \text{1 hydrogen} \times 1.01) = 137.34 + 2 \times (16.00 + 1.01) = 137.34 + 34.00 = 171.34 \, \text{g/mol}
\][/tex]
2. Convert the mass of barium hydroxide ([tex]\( \text{Ba(OH)}_2 \)[/tex]) to moles:
- Given mass of [tex]\( \text{Ba(OH)}_2 = 183.5 \, \text{g} \)[/tex]
- Molar mass of [tex]\( \text{Ba(OH)}_2 = 171.34 \, \text{g/mol} \)[/tex]
The number of moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] is calculated by:
[tex]\[
\text{Moles of Ba(OH)}_2 = \frac{183.5 \, \text{g}}{171.34 \, \text{g/mol}} \approx 1.071 \, \text{mol}
\][/tex]
3. Use the stoichiometric relationship from the balanced chemical equation:
- The balanced chemical equation is:
[tex]\[
2 \, \text{HBr} + \text{Ba(OH)}_2 \rightarrow \text{BaBr}_2 + 2 \, \text{H}_2\text{O}
\][/tex]
According to the balanced equation, 1 mole of [tex]\( \text{Ba(OH)}_2 \)[/tex] produces 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
4. Calculate the moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]) produced:
- We have [tex]\( 1.071 \, \text{moles of Ba(OH)}_2 \)[/tex].
- Therefore, the moles of water produced are:
[tex]\[
\text{Moles of H}_2\text{O} = 2 \times 1.071 \approx 2.142 \, \text{mol}
\][/tex]
Hence, 2.142 moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]) are formed when 183.5 g of [tex]\( \text{Ba(OH)}_2 \)[/tex] reacts with excess [tex]\( \text{HBr} \)[/tex].
