To determine how many moles of barium hydroxide, [tex]\( \text{Ba(OH)}_2 \)[/tex], are required to react with 117 grams of hydrogen bromide, [tex]\( \text{HBr} \)[/tex], we can follow a step-by-step process outlined below:
1. Calculate the molar mass of [tex]\( \text{HBr} \)[/tex]:
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
- Bromine (Br) has a molar mass of approximately 79.9 g/mol.
- Therefore, the molar mass of [tex]\( \text{HBr} \)[/tex] is:
[tex]\[
\text{Molar mass of HBr} = 1 \, \text{g/mol} + 79.9 \, \text{g/mol} = 80.9 \, \text{g/mol}
\][/tex]
2. Convert the given mass of [tex]\( \text{HBr} \)[/tex] to moles:
- Given mass of [tex]\( \text{HBr} \)[/tex] is 117 grams.
- Number of moles of [tex]\( \text{HBr} \)[/tex] can be calculated using the formula:
[tex]\[
\text{Moles of HBr} = \frac{\text{Mass of HBr}}{\text{Molar mass of HBr}} = \frac{117 \, \text{g}}{80.9 \, \text{g/mol}} \approx 1.446 \, \text{moles of HBr}
\][/tex]
3. Use the balanced chemical equation to determine the moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] required:
- The balanced equation is:
[tex]\[
2 \, \text{HBr} + \text{Ba(OH)}_2 \rightarrow \text{BaBr}_2 + 2 \, \text{H}_2\text{O}
\][/tex]
- According to the equation, 2 moles of [tex]\( \text{HBr} \)[/tex] react with 1 mole of [tex]\( \text{Ba(OH)}_2 \)[/tex].
4. Calculate the moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] required based on the moles of [tex]\( \text{HBr} \)[/tex] we have:
- From the mole ratio in the balanced equation (2:1), we need half as many moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] as we have moles of [tex]\( \text{HBr} \)[/tex]:
[tex]\[
\text{Moles of Ba(OH)}_2 = \frac{\text{Moles of HBr}}{2} = \frac{1.446}{2} \approx 0.723 \, \text{moles of Ba(OH)}_2
\][/tex]
Therefore, approximately 0.723 moles of barium hydroxide, [tex]\( \text{Ba(OH)}_2 \)[/tex], are required to react with 117 grams of hydrogen bromide, [tex]\( \text{HBr} \)[/tex].
