Question

Binary compounds of alkali metals and hydrogen react with water to liberate hydrogen gas. The hydrogen gas from the reaction of a sample of sodium hydride with an excess of water fills a volume of 0.510 L above the water. The temperature of the gas is 35 ∘C and the total pressure is 760 mmHg. Find the mass of H2 liberated and the mass of NaH that reacted.

Answer 1

Answer: The mass of $H_{2}$ liberated is 0.0383 g and the mass of NaH that reacted 0.455 g.

Explanation:

The given data is as follows.

       Volume = 0.501 L,

 Temperature of gas = $35^{o}C$ = (35 + 273) K = 308 K,

 Total pressure = 760 mm Hg

  Vapor pressure of water at $35^{o}C$ = 422 mm Hg

  Partial pressure of $H_{2}$ = (760 - 422) mm Hg

                                     = 717.8 mm Hg

The chemical equation is as follows.

       $NaH + H_{2}O \rightarrow NaOH + H_{2}$

Using ideal gas equation, we will find the moles of hydrogen gas as follows.

             PV = nRT

         n = $\frac{PV}{RT}$

As we know,

               1 atm = 760 mm Hg

      717.8 mm Hg = $717.8 \times \frac{1 atm}{760 mm Hg}$    

                             = 0.944 atm

Putting the given values into the above formula we will calculate the number of moles as follows.

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

                       = $\frac{0.944 atm \times 0.510 L}{0.08206 Latm/mol K \times 308 K}$

                       = 0.0190 moles

Now, we will find the mass of $H_{2}$ as follows.

       Mass = Moles × Molar mass

                 = 0.0190 moles × 2.01588 g/mol

                 = 0.0383 g

Therefore, mass of hydrogen is 0.0383 g.

As the ratio between $H_{2}$ and NaH is 1:1. So, we will calculate the moles of NaH as follows.

             Mass = Moles × Molar mass

                       = 0.0190 moles × 23.99771 g/mol

                       = 0.455 g

Therefore, mass of NaH is 0.455 g.