Question

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

Answer 1

Answer: The mass or zinc reacted is 0.624 grams.

Explanation:

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm    (Conversion factor:  1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

$p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm$

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L    (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = $30^oC=[30+273]K=303K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol$

The chemical equation for the reaction of zinc and hydrochloric acid follows:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, $9.55\times 10^{-3}mol$ of hydrogen gas is produced from = $\frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol$ of zinc metal

To calculate the mass of zinc metal, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of zinc = 65.38 g/mol

Moles of zinc = $9.55\times 10^{-3}$ moles

Putting values in above equation, we get:

$9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g$

Hence, the mass or zinc reacted is 0.624 grams.