Apiece of sodium metal reacts completely with water to produce sodium hydroxide and hydrogen gas. the hydrogen gas generated is collected over water at 25.0 degrees celsius. the volume of the gas is 246 ml measured at 1.00 atm. calculate the number of grams of sodium used in the reaction (vapor pressure of water at 25 degrees celsius = 0.0313 atm).

We get the pressure of the hydrogen gas from the difference between the measured pressure and the vapor pressure of water:
     total pressure = Pressure of H2 + Vapor Pressure of H2O
     1.00 atm = Pressure of H2 + 0.0313 atm
     Pressure of H2 = 1.00 atm - 0.0313 atm = 0.9687 atm

From the ideal gas law,
     PV = nRT
we can calculate for the number of moles of H2 as
     n = PV/RT = (0.9687 atm)(0.246L) / (0.08206 L·atm/mol·K)(298.15 K)
        = 0.00974 mol H2
where 
     V = 246 mL (1 L / 1000 mL) = 0.246 L
     T = 25 degrees Celsius + 273.15 = 298.15 K 

We use the mole ratio of Na and H2 from the reaction of sodium metal with water as shown in the equation 
     2Na(s) + 2H2O(l) → 2 NaOH(aq) + H2(g)
and the molar mass of sodium Na to get the mass of sodium used in the reaction:
     mass of Na = 0.00974 mol H2 (2 mol Na /1 mol H2)(22.99 g Na/1 mol Na)
                        = 0.448 grams of sodium

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