Certainly! Let's go through the problem step-by-step to determine the volume of [tex]\( H_2 \)[/tex] gas formed at STP when 3.82 g of Al reacts with excess NaOH. Here's a structured approach:
1. Write the Balanced Chemical Equation:
[tex]\[
2 \text{NaOH} (s) + 2 \text{Al} (s) + 6 \text{H}_2\text{O} (l) \rightarrow 2 \text{NaAl}(\text{OH})_4 (s) + 3 \text{H}_2 (g)
\][/tex]
2. Identify the Molar Mass of Aluminum (Al):
[tex]\[
\text{Molar mass of Al} = 26.98 \text{ g/mol}
\][/tex]
3. Calculate the Moles of Aluminum (Al):
[tex]\[
\text{Given mass of Al} = 3.82 \text{ g}
\][/tex]
Using the molar mass, we can find the moles of Al:
[tex]\[
\text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{3.82 \text{ g}}{26.98 \text{ g/mol}} \approx 0.141586 \text{ mol}
\][/tex]
4. Determine the Moles of [tex]\( H_2 \)[/tex] Gas Produced:
From the balanced equation, 2 moles of Al produce 3 moles of [tex]\( H_2 \)[/tex] gas. Hence, we can set up the relation:
[tex]\[
\text{Moles of } H_2 = \frac{3}{2} \times \text{Moles of Al} = \frac{3}{2} \times 0.141586 \text{ mol} \approx 0.212380 \text{ mol}
\][/tex]
5. Calculate the Volume of [tex]\( H_2 \)[/tex] Gas at STP:
At Standard Temperature and Pressure (STP), 1 mole of an ideal gas occupies 22.4 L. Using this information, we can calculate the volume of the [tex]\( H_2 \)[/tex] gas:
[tex]\[
\text{Volume of } H_2 = \text{Moles of } H_2 \times \text{Volume at STP} = 0.212380 \text{ mol} \times 22.4 \text{ L/mol} \approx 4.7573 \text{ L}
\][/tex]
6. Final Answer:
Hence, the volume of [tex]\( H_2 \)[/tex] gas formed at STP when 3.82 g of Al reacts with excess NaOH is approximately [tex]\( 4.76 \text{ L} \)[/tex].
So, the correct answer is:
[tex]\[
\boxed{4.76 \text{ L}}
\][/tex]
