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How many moles of sulfuric acid [tex](H_2SO_4)[/tex] are needed to react completely with 6.8 moles of lithium hydroxide [tex](LiOH)[/tex]?

[tex]2 \, \text{LiOH} + H_2SO_4 \rightarrow Li_2SO_4 + 2 \, \text{H}_2O[/tex]

A. [tex]3.4 \, \text{mol} \, H_2SO_4[/tex]
B. [tex]6.8 \, \text{mol} \, H_2SO_4[/tex]
C. [tex]10.2 \, \text{mol} \, H_2SO_4[/tex]
D. [tex]13.6 \, \text{mol} \, H_2SO_4[/tex]



Answer :

To determine how many moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] are needed to react completely with 6.8 moles of [tex]\( \text{LiOH} \)[/tex], we can follow these steps:

1. Write the 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]

2. Determine the mole ratio from the balanced equation:

From the balanced chemical equation, we see that 2 moles of [tex]\( \text{LiOH} \)[/tex] react with 1 mole of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]. This gives us the mole ratio:
[tex]\[ \frac{2 \text{ moles LiOH}}{1 \text{ mole H}_2\text{SO}_4} \][/tex]

3. Calculate the moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] required:

Given that we have 6.8 moles of [tex]\( \text{LiOH} \)[/tex], we can use the mole ratio to find the amount of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] needed.

Since 2 moles of [tex]\( \text{LiOH} \)[/tex] react with 1 mole of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{\text{Moles of } \text{LiOH}}{2} = \frac{6.8 \text{ moles LiOH}}{2} \][/tex]

4. Perform the division:

[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{6.8}{2} = 3.4 \][/tex]

Therefore, 3.4 moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] are needed to react completely with 6.8 moles of [tex]\( \text{LiOH} \)[/tex].

So, the correct answer is:
[tex]\[ 3.4 \text{ mol } \text{H}_2\text{SO}_4 \][/tex]