For sodium reacting with water we have the equation:

2Na (s) + 2H2O -> 2NaOH + H2

One mole of sodium reacts with one mole of water to give one mole of NaOH and half a mole of H2.

The molar mass of sodium is 22.98

48.7 grams of sodium are equivalent to 2.11 moles.

As the reaction yields 0.5 mole of H2 per mole of sodium, for 2.11 moles of sodium we have 1.06 moles of hydrogen.

One mole of hydrogen has a number of molecules equal to the Avogadro Constant or 6.022* 10^23. 1.06 moles of hydrogen is equal to 6.022* 10^23* 1.06 = 6.38*10^23 molecules.

**Therefore the number of molecules of hydrogen that 48.7 g of sodium added to water yields is 6.38*10^23.**

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