Question

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose 226 mL of hydrogen gas is collected at 30.°C and has a total pressure of 1.018 atm by this process. What is the partial pressure of hydrogen gas in the sample? (The vapor pressure of water is 32 torr at 30°C.) atm

Answer 1

Explanation:

The given data is as follows.

          Pressure of moist hydrogen gas = 1.018 atm

          Pressure of water vapor = $32 torr \times \frac{1 atm}{760 torr}$    

                                                   = 0.0421 atm

Formula to calculate the pressure of dry hydrogen gas is as follows.

       Pressure of $H_{2}$ gas = $P_{moist H_{2}} - P_{H_{2}O}$

Now, putting the given values into the above formula as follows.

   Pressure of $H_{2}$ gas = $P_{moist H_{2}} - P_{H_{2}O}$

                             = 1.018 atm - 0.0421 atm

                             = 0.976 atm

Therefore, we can conclude that partial pressure of hydrogen gas in the given sample is 0.976 atm.  

Answer 2

Answer:

0.582 g

Explanation:

The vapor pressure of the water = 32 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 32 / 760 atm = 0.0421 atm

Total pressure = 1.018 atm

Pressure of hydrogen gas = 1.018 - 0.0421 atm = 0.9759 atm

Temperature = 30 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (30 + 273.15) K = 303.15 K  

Volume = 226 mL = 0.226 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9759 atm × 0.226 L = n × 0.0821 L.atm/K.mol × 303.15 K  

⇒n = 0.0089 moles

According to the given reaction,

$Zn_{(s)} + 2 HCl_{(aq)}\rightarrow ZnCl_2_{(aq)} + H_2_{(g)}$

1 mole of hydrogen gas is produced from 1 mole of Zn

0.0089 mole of hydrogen gas is produced from 0.0089 mole of Zn

Thus, Moles of zinc = 0.0089 moles

Molar mass of zinc = 65.39 g/mol

Mass= Moles * Molar mass = 0.0089 * 65.39 g = 0.582 g