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