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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.

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The mass of NaH reacted is 0.456 g.

We have the following information from the question;

Volume of hydrogen (V)= 0.510 L

Temperature (T)=  35 ∘C + 273 = 308 K

Pressure of hydrogen(P) =  760 mmHg - 31.82 mm Hg = 728.18  mm Hg or 0.96 atm

Gas constant (R) = 0.082 atm LK-1mol-1

Number of moles(n) = ?

From;

PV= nRT

n = PV/RT

n = 0.96 atm ×  0.510 L/0.082 atm LK-1mol-1  ×  308 K

n = 0.019 moles

Mass of H2 liberated = 0.019 moles × 2 g/mol =  0.038 g of H2

The equation of the reaction is;

NaH(s) + H2O(l) ----->NaOH(aq) + H2(g)

Since  moles of NaH yields 1 mole of H2

0.019 moles of NaH yields  0.019 moles of H2

Mass of NaH =  0.019 moles × 24 g/mol = 0.456 g

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