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 0.475 L above the water. The temperature of the gas is 35 ∘C and the total pressure is 755 mmHg.

Find the mass of H2 liberated and the mass of NaH that reacted.

Answer 1

• 0.03733 g H₂

• 0.4480 g NaH

Step-by-step explanation:

The reaction that takes place is

• NaH(s) + H₂O(l) → NaOH(aq) + H₂(g)

To find the mass of H₂ liberated we use PV=nRT:

1. 755 mmHg ⇒755/760 = 0.993 atm
2. 35°C ⇒ 35 + 273.16 = 308.16 K

• 0.993 atm * 0.475 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 308.16 K

• n = 0.01867 mol H₂

• Mass H₂ = 0.01867 mol * 2g/mol = 0.03733 g

Then we use the moles of H₂ formed by the reaction to calculate the moles of NaH that reacted, and finally the mass using its molecular weight:

• 0.01867 mol H₂ *
  `(1molNaH)/(1molH_(2)) *(24g)/(1molNaH)` = 0.4480 g NaH