To determine how many moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] are required to completely neutralize 8.762 moles of [tex]\( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \)[/tex], we need to balance the chemical equation first and then use the stoichiometric relationship.
1. Balance the chemical equation:
The given unbalanced equation is:
[tex]\[
\text{H}_3\text{C}_6\text{H}_5\text{O}_7(aq) + \text{Ba(OH)}_2(aq) \rightarrow \text{H}_2\text{O}(l) + \text{Ba}_3(\text{C}_6\text{H}_5\text{O}_7)_2(aq)
\][/tex]
To balance the equation, we need to match the number of atoms of each element on the reactants side with those on the products side.
- Balance the Ba atoms:
[tex]\[
\text{Ba(OH)}_2(aq) \rightarrow \text{Ba}_3(\text{C}_6\text{H}_5\text{O}_7)_2(aq)
\][/tex]
Since there are 3 Ba atoms on the product side, we need 3 Ba atoms on the reactants side:
[tex]\[
\text{3 Ba(OH)}_2(aq)
\][/tex]
- Balance the Citric Acid ([tex]\( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \)[/tex]):
[tex]\[
2 \text{H}_3\text{C}_6\text{H}_5\text{O}_7(aq)
\][/tex]
- Now, let's balance the Hydrogen (H) and Oxygen (O) atoms:
The total H atoms in the reactants:
[tex]\[
2 \times 3 (\text{H}) + 3 \times 2 (\text{OH}) = 6 \text{H} + 6 \text{OH}
\][/tex]
This breaks into 6 molecules of water ([tex]\(\text{H}_2\text{O}\)[/tex]):
[tex]\[
6 \text{H}_2\text{O}(l)
\][/tex]
The balanced equation is therefore:
[tex]\[
2 \text{H}_3\text{C}_6\text{H}_5\text{O}_7(aq) + 3 \text{Ba(OH)}_2(aq) \rightarrow 6 \text{H}_2\text{O}(l) + \text{Ba}_3(\text{C}_6\text{H}_5\text{O}_7)_2(aq)
\][/tex]
2. Determine the moles of [tex]\(\text{Ba(OH)}_2\)[/tex] required:
From the balanced equation, we see that 2 moles of [tex]\(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\)[/tex] require 3 moles of [tex]\(\text{Ba(OH)}_2\)[/tex]:
The mole ratio of [tex]\(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\)[/tex] to [tex]\(\text{Ba(OH)}_2\)[/tex] is:
[tex]\[
\frac{3}{2}
\][/tex]
Given [tex]\( 8.762 \)[/tex] moles of [tex]\(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\)[/tex]:
[tex]\[
\text{Moles of } \text{Ba(OH)}_2 = 8.762 \times \frac{3}{2} = 13.143 \text{ moles}
\][/tex]
Hence, 13.143 moles of [tex]\(\text{Ba(OH)}_2\)[/tex] are required to completely neutralize 8.762 moles of [tex]\(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\)[/tex].
