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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