Question

According to the equation 2Na + 2H2O mc012-1.jpg 2NaOH+H2, what mass of Na is required to yield 22.4 L of H2 at STP? (The atomic mass of Na is 22.99 u.)

Answer 1

$2Na_{(s)} + 2H _{2} O_{(l)} ------\ \textgreater \ 2NaOH_{(aq)} + H_{2}_{(g)}$

moles of hydrogen in reaction  =  $\frac{Volume}{Moles at STP}$
                                        
                                                 = $\frac{22.4 L}{22.5 L / mol}$
         
                                                 = 0.996 mol

ratio of $H_{2}$  :  Na  is  1 : 2
∴ if mol of $H_{2}$ = 0.996 mol

 then mol of Na      =   (0.996 mol * 2)
                              = 1.99 mol

Mass of Na  =  molar mass * mol
                    =  (22.99 g / mol) * (1.99 mol)
                    =  45.78 g

Note: 
   1) Molar Mass is the mass of an element measured in grams.

   2) According to Avogadro's Law, the same volume of different gases at the same condition of Temperature and Pressure contain the same number of particles.  From this law we know that at STP, one mole of gas occupies 22.4 L of volume (molar volume).

3) By calculating the mole of one specie in a reaction, one can use mole ratio based on the stoichimectric values (values used to balance the equation) given to the species upon balancing the equation  to find the moles of another species.

Answer 2

45.78 just took the test so 46.0 round up