Question

A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25 ∘ C. What is the percent yield of hydrogen for this reaction?

Answer 1

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 = $\frac{PV}{RT}$

                 = $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

                  = 0.0052 mol

Also,  No. of moles = $\frac{mass}{\text{molar mass}}$

               0.0052 mol = $\frac{mass}{2 g/mol}$

                  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 $25^{o}C$

                $P_{\text{water vapor}}$ = 24 mm Hg

                                = $\frac{24}{760}$ atm

                                = 0.03158 atm

Now,   P = $\frac{756}{760} - 0.03158$

              = 0.963 atm

Hence,   n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

                 = 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

                         = 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

      Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

                              = $\frac{0.01013 g}{0.0104 g} \times 100$            

                              = 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.