To solve this problem, we need to use stoichiometry based on the given balanced chemical equation:
[tex]\[ 2 \text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
From the balanced equation, we see that 2 moles of lithium hydroxide (LiOH) react with 1 mole of sulfuric acid (H_2SO_4).
Let's break this down step-by-step:
1. Identify the mole ratio from the chemical equation:
According to the equation, 2 moles of LiOH react with 1 mole of H_2SO_4. This gives us a stoichiometric ratio of:
[tex]\[ \frac{\text{moles of LiOH}}{\text{moles of H}_2\text{SO}_4} = \frac{2}{1} \][/tex]
2. Determine the moles of sulfuric acid needed:
Here, we are given that we have 6.8 moles of LiOH. Using the mole ratio from the equation, we can find the moles of H_2SO_4 needed to react completely with 6.8 moles of LiOH.
[tex]\[
\text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of LiOH}}{2} = \frac{6.8}{2}
\][/tex]
3. Calculate the moles of sulfuric acid needed:
[tex]\[
\text{Moles of H}_2\text{SO}_4 = 3.4
\][/tex]
Therefore, 3.4 moles of sulfuric acid ([tex]\( \text{H}_2\text{SO}_4 \)[/tex]) are required to react completely with 6.8 moles of lithium hydroxide ([tex]\( \text{LiOH} \)[/tex]).
Thus, the correct answer is:
[tex]\[ \boxed{3.4 \text{ mol H}_2\text{SO}_4} \][/tex]
