Question

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.

$\mathbf{Zn}\left(\mathbf{s}\right)\mathbf{+}\mathbf{2}\mathbf{HCl}\left(\mathbf{aq}\right)\to {\mathbf{ZnCl}}_{\mathbf{2}}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(\mathbf{g}\right)$

Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose 240. mL of hydrogen gas is collected at 30.^oC and has a total pressure of 1.032 atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at 30^oC.)

Short answer

The partial pressure of hydrogen gas in the sample is 0.990 atm.

The mass of zinc reacted to generate this much amount of hydrogen gas is 0.625 g.

Step-by-step solution

Step1:

The partial pressure of hydrogen gas in the sample can be determined as follows:

${\text{P}}_{\text{total}}{\text{=P}}_{{\text{H}}_{\text{2}}}{\text{+P}}_{{\text{H}}_{\text{2}}\text{O}}$

Rearrange the above equation to ${\text{P}}_{{\text{H}}_{\text{2}}}$

$$\begin{gathered} {\text{P}}_{{\text{H}}_{\text{2}}}{\text{=P}}_{\text{total}} \text{-} {\text{P}}_{{\text{H}}_{\text{2}}\text{O}} \\ \text{= 1.032} \text{atm} \text{-32torr×}\frac{1atm}{760torr} \\ \text{=0.990atm} \end{gathered}$$

Therefore, the partial pressure of hydrogen gas in the sample is 0.990 atm.

The number of moles of hydrogen

The moles of hydrogen gas in the sample is obtained by using the ideal gas equation.

$\text{n =}\frac{\text{PV}}{\text{RT}}$

Here,

n is the number of moles.

P is the pressure.

V is the volume.

T is the temperature.

R is the gas constant $\text{0.08206} \text{L} \text{atm/mol} \text{K}$.

$$\begin{gathered} n=\frac{0.990atm\times 0.240L}{0.08206atm/molK\times 303K} \\ =9.56\times {10}^{-3}mol \end{gathered}$$

Hence, the moles of hydrogen gas in the sample is$9.56\times {10}^{-3}mol$.

Mass of Zinc

According to the given balanced equation, one mole of zinc reacts with two moles of hydrochloric acid, giving one mole of zinc chloride and one mole of hydrogen gas. Thus, the stoichiometric ratio of zinc to hydrogen is $1:1$.

The mass of zinc reacted to produce $9.56\times {10}^{-3}mol$ of hydrogen gas can be calculated by using the below expression.

$9.56\times {10}^{-3}mol{H}_2\times \frac{ 1\text{mol} \text{Zn}}{\text{1} \text{mol} {\text{H}}_{\text{2}}}\times \frac{\text{65.38} \text{g} \text{Zn}}{\text{1} \text{mol} \text{Zn}}=0.625gZn$

Therefore, the mass of zinc reacted to generate this amount of hydrogen gas is 0.625 g.