To determine how many moles of [tex]\( H_2SO_4 \)[/tex] are needed to completely neutralize 0.0164 moles of [tex]\( KOH \)[/tex], we start by considering the balanced chemical equation for the neutralization reaction:
[tex]\[ H_2SO_4 + 2 KOH \rightarrow K_2SO_4 + 2 H_2O \][/tex]
This equation indicates that one mole of [tex]\( H_2SO_4 \)[/tex] reacts with two moles of [tex]\( KOH \)[/tex]. From the stoichiometry of the equation, for every 1 mole of [tex]\( H_2SO_4 \)[/tex] needed, 2 moles of [tex]\( KOH \)[/tex] are required.
Given that we have 0.0164 moles of [tex]\( KOH \)[/tex], we need to find out how many moles of [tex]\( H_2SO_4 \)[/tex] will be necessary to fully neutralize this amount. We can set up the ratio:
[tex]\[ \frac{1 \text{ mole } H_2SO_4}{2 \text{ moles } KOH} = \frac{x \text{ moles } H_2SO_4}{0.0164 \text{ moles } KOH} \][/tex]
Solving for [tex]\( x \)[/tex] (the moles of [tex]\( H_2SO_4 \)[/tex]):
[tex]\[ x = \frac{0.0164 \text{ moles } KOH \times 1 \text{ mole } H_2SO_4}{2 \text{ moles } KOH} \][/tex]
[tex]\[ x = \frac{0.0164}{2} \][/tex]
[tex]\[ x = 0.0082 \text{ moles of } H_2SO_4 \][/tex]
Therefore, 0.0082 moles of [tex]\( H_2SO_4 \)[/tex] are needed to completely neutralize 0.0164 moles of [tex]\( KOH \)[/tex].
