Sure, let's solve this step-by-step:
Given Data:
- Mass of Zinc ([tex]\(Zn\)[/tex]): [tex]\(25.0 \, \text{g}\)[/tex]
- Molar mass of [tex]\(Zn\)[/tex]: [tex]\(65.38 \, \text{g/mol}\)[/tex]
- Temperature: [tex]\(25^{\circ} \text{C}\)[/tex]
- Pressure: [tex]\(854 \, \text{mmHg}\)[/tex]
- Gas constant ([tex]\(R\)[/tex]): [tex]\(0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\)[/tex]
1. Convert Temperature to Kelvin:
The temperature in Kelvin ([tex]\(K\)[/tex]) is:
[tex]\[
T_K = 25 + 273.15 = 298.15 \, \text{K}
\][/tex]
2. Convert Pressure to Atmospheres:
The pressure in atmospheres ([tex]\(\text{atm}\)[/tex]) can be calculated as:
[tex]\[
P_{\text{atm}} = \frac{854 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 1.1236842105263158 \, \text{atm}
\][/tex]
3. Calculate the Moles of Zinc (Zn):
Using the molar mass of Zinc:
[tex]\[
\text{Moles of } Zn = \frac{25.0 \, \text{g}}{65.38 \, \text{g/mol}} = 0.38237993270113185 \, \text{mol}
\][/tex]
4. Determine the Moles of Hydrogen Gas ([tex]\(H_2\)[/tex]) Produced:
From the balanced chemical equation [tex]\(Zn (s) + 2 HCl (aq) \rightarrow H_2 (g) + ZnCl_2 (aq)\)[/tex], 1 mole of [tex]\(Zn\)[/tex] produces 1 mole of [tex]\(H_2\)[/tex].
Therefore:
[tex]\[
\text{Moles of } H_2 = 0.38237993270113185 \, \text{mol}
\][/tex]
5. Calculate the Volume of [tex]\(H_2\)[/tex] Gas:
Using the Ideal Gas Law, [tex]\( PV = nRT \)[/tex], we need to solve for the volume ([tex]\(V\)[/tex]):
[tex]\[
V = \frac{nRT}{P}
\][/tex]
Where:
- [tex]\(n = 0.38237993270113185 \, \text{mol}\)[/tex]
- [tex]\(R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\)[/tex]
- [tex]\(T = 298.15 \, \text{K}\)[/tex]
- [tex]\(P = 1.1236842105263158 \, \text{atm}\)[/tex]
Plugging in these values:
[tex]\[
V = \frac{0.38237993270113185 \, \text{mol} \times 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \times 298.15 \, \text{K}}{1.1236842105263158 \, \text{atm}} = 8.329688963028607 \, \text{L}
\][/tex]
So, when [tex]\(25.0 \, \text{g}\)[/tex] of [tex]\(Zn\)[/tex] reacts, the volume of [tex]\(H_2\)[/tex] gas formed is approximately [tex]\(8.33 \, \text{L}\)[/tex].
From the given options, the closest match is:
- [tex]\(8.32 \, \text{L}\)[/tex]
Thus, the correct answer is [tex]\(8.32 \, \text{L}\)[/tex].
