# Question #04e6f

Dec 7, 2014

The answer is $0.13 g r a m s$.

$Z n \left(s\right) + 2 H C l \left(a q\right) \to Z n C {L}_{2} \left(a q\right) + {H}_{2} \left(g\right)$

Now, the gas collected in t he $36.6 m L$ tube is hydrogen mixed with water vapor. In order to get the amount of hydrogen (${H}_{2}$), we must get its partial pressure in the mixture by using Dalton's law.

So, the total pressure collected over the water is equal to the partial pressure of water and the partial pressure of ${H}_{2}$.

${P}_{c o l l e c t e d} = {P}_{{H}_{2}} + {P}_{w a t e r}$, which means

${P}_{{H}_{2}} = {P}_{c o l l e c t e d} - {P}_{w a t e r} = 890.0 - 17.5 = 872.5 m m H g$.

Using the ideal gas law, $P V = n R T$, we can determine the number of moles of ${H}_{2}$ produced in the reaction

${n}_{{H}_{2}} = \frac{P V}{R T} = \frac{\left(\frac{872.5}{760}\right) \cdot 36.6 \cdot {10}^{- 3}}{0.082 \cdot \left(20 + 273.15\right)} = 0.002 m o l e s$ (notice the conversions of pressure, volume and temperature to atm, L, and K, respectively);

From the balanced equation we can see that we have a $1 : 1$ mole ratio between $Z n$ and ${H}_{2}$, which means that the number of moles of $Z n$ is equal to

${n}_{Z n} = 0.002$ moles

Therefore, the mass of $Z n$ (knowing that its molar mass is $65.4 \frac{g}{m o l}$), is equal to

${m}_{Z n} = {n}_{Z n} \cdot 65.4 \frac{g}{m o l} = 0.004 m o l e s \cdot 65.4 \frac{g}{m o l} = 0.13 g$