Question

Reaction of 0.028 g of magnesium with excess hydrochloric acid generated 31.0 mL of hydrogen gas. The gas was collected by water displacement at 22 degree C. The barometric pressure in the lab was 746 mm Hg. The vapor pressure of water at 22 degree C is 19.8 mm Hg. Use Dalton's law to calculate the partial pressure of hydrogen gas in the gas-collecting tube. Use the combined gas law to calculate the volume of hydrogen at STP. What is the theoretical number of moles of hydrogen that can be produced from 0.028g of Mg?

Answer 1

Explanation:

(a)   It is given that magnesium is reacted with hydrochloric acid and the hydrogen evolved is collected at top. This means that hydrochloric acid will be present in a solution (HCl + Water) and the solvent will be water.

Due to evaporation some amount of water will have evaporated and would be present in vapor phase. Therefore, when the reaction occurs only hydrogen will not be present in vapor phase but, will be accompanied by water vapors as well .

Hence, Dalton's law the total pressure of the system will be sum of pressure exerted by hydrogen gas and pressure exerted by water vapors .

Let us assume that the partial pressure of hydrogen gas be "$P_H_{2}$"

And, the partial pressure of water will be nothing but the vapor pressure of water,

Vapor pressure of water = $P_{water}$

                                         = 19.8 mm Hg

Total pressure of the system = 746 mm Hg

Total pressure = $P_H_{2} + P_{water}$

               746 = $P_H_{2}$ + 19.8

or,               $P_H_{2}$ = 746-19.8

                                = 726.2 mm Hg

Hence, partial pressure of hydrogen gas is 726.2 mm Hg.

(b) To calculate volume at STP, we will first calculate at $22^{o}C$ and 726.2 mm Hg and than convert it to STP conditions.

Therefore, to calculate volume at $22^{o}C$ and 726.2 mm Hg we will make use of ideal gas law  as follows.

           P = 726.2 mm Hg

             = $\frac{726.2}{760}$

             = 0.955 atm

          T = $22^{o}C$

              = 22+273.15 = 295.15 K

         V = 31 ml = $31 \times 10^{-3}$ Litre

According to the ideal gas law ,

              PV = nRT

where,      P = pressure of the system ,  

                 V = volume of the gas

                 N = number of moles

                R = 0.0821 liter atm/mole K

                T = Temperature

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

   $0.955 \times 31 \times 10^{-3} = N \times 0.0821 \times 295.15$

                      N = $1.222 \times 10^{-3}$ moles

Now, the moles of hydrogen won't change. Therefore, let us calculate  volume at STP of $1.222 \times 10^{-3}$ moles of hydrogen.

Now, at STP T = 273.15 K , P = 1 atm  and N = $1.222 \times 10^{-3}$ moles

    $1 \times V = 1.222 \times 10^{-3} \times 0.0821 \times 273.15 K$

       V = 0.027398 Litre

         = $0.027398 \times 1000$          (as 1 L = 1000 ml)

         = 27.398 ml

Therefore, volume of hydrogen at STP is 27.398 ml .

(c)  Now, we can write the the reaction for this case as follows.

         $Mg + 2HCl \rightarrow MgCl_{2} + H_{2}$

As, weight of magnesium = 0.028 grams

      Molar mass of magnesium = 24.3 grams/mole

   Number of moles of magnesium = $\frac{mass}{\text{molar mass}}$

                                             = $\frac{0.028}{24.3}$

                                             = $1.15226 \times 10^{-3}$ moles

Since, it can be seen from the reaction  that

    1 mole of Magnesium = 1 mole of hydrogen

and, moles of hydrogen = $1.15226 \times 10^{-3}$ moles

                                        = 0.001523 moles

Hence, theoretical number of moles of hydrogen that can be produced from 0.028 grams of Mg is 0.001523 moles