Respuesta :
Explanation:
As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.
So, using ideal gas equation we will calculate the number of moles as follows.
PV = nRT
or, n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}[/tex]
= 0.0052 mol
Also, No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
0.0052 mol = [tex]\frac{mass}{2 g/mol}[/tex]
mass = 0.0104 g
As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at [tex]25^{o}C[/tex]
[tex]P_{\text{water vapor}}[/tex] = 24 mm Hg
= [tex]\frac{24}{760}[/tex] atm
= 0.03158 atm
Now, P = [tex]\frac{756}{760} - 0.03158[/tex]
= 0.963 atm
Hence, n = [tex]\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}[/tex]
= 0.0056 mol
So, mass of [tex]H_{2}[/tex] = 0.0056 mol × 2
= 0.01013 g (actual yield)
Therefore, calculate the percentage yield as follows.
Percent yield = [tex]\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100[/tex]
= [tex]\frac{0.01013 g}{0.0104 g} \times 100[/tex]
= 97.49%
Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.
