Question

3、$\ce{h_{2}so_{4} + 2 naoh \
ightarrow 2 h_{2}o + na_{2}so_{4}}$ how many moles of naoh are needed to neutralize $5.20 \times 10^{24}$ particles of $\ce{h_{2}so_{4}}$?

Explanation:

Step1: Find moles of $H_2SO_4$

Using Avogadro's number ($N_A = 6.022\times10^{23}$ particles/mol), moles of $H_2SO_4$ ($n_{H_2SO_4}$) is calculated as:
$n_{H_2SO_4} = \frac{\text{Number of particles}}{N_A} = \frac{5.20\times10^{24}}{6.022\times10^{23}}$

Step2: Use stoichiometry

From the reaction: $1$ mol $H_2SO_4$ reacts with $2$ mol $NaOH$.
So moles of $NaOH$ ($n_{NaOH}$) = $2 \times n_{H_2SO_4}$

Calculation for Step1:

$n_{H_2SO_4} = \frac{5.20\times10^{24}}{6.022\times10^{23}} \approx 8.635$ mol

Calculation for Step2:

$n_{NaOH} = 2 \times 8.635 \approx 17.27$ mol

Answer:

Approximately $\boldsymbol{17.3}$ moles (or more precisely ~17.27 moles) of $NaOH$ are needed.