Binary compounds of alkali metals and hydrogen react with water to produce h2(g). the h2 from the reaction of a sample of nah 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 755 mmhg .

Mass of H2 liberated = 0.039g;

Mass of NaH that reacted = 0.468g

Note: This is the complete question (Binary compounds of alkali metals and hydrogen react with water to liberate H2(g). The H2 from the reaction of a sample of NaH 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 755mmHg . Find the mass of H2 liberated and the mass of NaH that reacted.)

Explanation:

The equation of the reaction is NaH + H2O ---> NaOH + H2O

molar mass of NaH = 24g, molar mass of H2 = 2g

First, The volume, V1, of the gas produced at standard conditions (s.t.p.) is determined using the general gas equation; P1V1/T1 = P2V2/T2

Making V1 subject of the formula; V1 = P2V2T1/P1T2

P1 = 760mHg, P2 = 755mmHg, V1 = ?, V2 = 0.5L, T1 = 0∘C  = 273K, T2 = 35 ∘C = 308K

V1 = 755*0.5*273/(760*308) = 0.44L

1 mole or 22.4L of H2 has a mass of 2g;

mass of 0.44L of H2 = 2*0.44/22.4 = 0.039g

From the equation of the reaction, 2g of H2 is produced by 24g of NaH at s.t. p.

0.039g of H2 will be produced by 0.039*24/2 = 0.468g of NaH