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 0.500 L above the water. The temperature of the gas is 35 ∘C and the total pressure is 750 mmHg .

Part A

Find the mass of hydrogen gas liberated.

Part B

Find the mass of sodium hydryide that reacted.

Answer 1

Given

Volume of H ₂ gas = 0.500 L

Temperature = 35.0 ⁰ C = 308.15 K

Total pressure of gas = 750 mm Hg

According to Dalton's law, When gas is collected over water then the total pressure of a gas is the sum of the partial pressure of a gas and partial pressure of water.

i e P total = P gas + P water vapor.

In this case, P total = P H₂ gas + P water vapor.

Pressure of Dry H₂ gas = P total - P water vapor

Pressure of Dry H₂ gas = 750 mm Hg - 42.23 mm Hg

Pressure of Dry H₂ gas =707.77 mm Hg

We have relation, 1 atm = 760 mm Hg.

Hence, Pressure of Dry H₂ gas = 707.77 mm Hg $\times$ (1 atm / 760 mm Hg) = 0.931 atm

We have relation, P V = n R T

Where P is pressure of a gas, V is a volume of a gas , n is no. of moles of gas and T is temperature of gas.

Therefore, number of moles of H₂ gas = P V / R T

number of moles of H₂ gas = 0.931 atm $\times$ 0.500 L / 0.082057 L atm / mol K $\times$ 308.15 K

number of moles of H₂ gas = 0.0184 mol

We have, Number of moles = Mass / Molar mass

Hence, Mass of  H₂ gas = No of moles $\times$ Molar mass

= 0.0184 mol $\times$ 2.0158 g / mol

= 0.0371 g

ANSWER : Mass of  H₂ gas = 0.0371 g

Consider reaction , NaH (aq) + H₂O (l) NaOH (aq) +  H_{2 (g)}

From reaction, 1 mol NaH $\equiv$ 1 mol   H_{2 (g)}

Therefore, 0.0184 mol NaH $\equiv$ 0.0184 mol   H_{2 (g)}

Molar Mass of NaH = 22.99 + 1.0079 = 24.00 g / mol

Mass of NaH = No of moles x Molar mass

= 0.0184 mol x (24.00 g /1 mol)

= 0.442 g

ANSWER: Mass of NaH = 0.442 g