What mass of sodium hydroxide is formed when 2.3 g of sodium reacts with water?

When sodium interacts with water, the reaction occurs:

2Na + 2H2O = 2NaOH + H2.

As can be seen, 2 mol of Na enters this reaction and 2 mol of NaOH is formed. Therefore, it can be argued that when 1 mol of Na is consumed, 1 mol of NaOH is formed. It follows that:

From 23 g of Na, 40 g of NaOH are formed (according to the formula m = n * Mr).

Let’s make the following proportion:

2.3 g – x g

23 g – 40 g

From here we find that x = 2.3 * 40/23 = 4 g (the mass of NaOH formed by the interaction of 2.3 g of Na with H2O).