What amount of sodium hydroxide is formed during the interaction 72g. water with sodium?

To solve the problem, it is necessary to draw up an equation, select the coefficients:
2Na + 2H2O = 2NaOH + H2 – redox reaction, sodium hydroxide obtained;
Let’s make calculations using the formulas:
M (H2O) = 18 g / mol;
M (NaOH) = 39.9 g / mol.
Let us determine the amount of moles of water, if the mass is known:
Y (H2O) = m / M = 72/18 = 4 mol.
Let’s make a proportion according to the equation:
4 mol (H2O) – X mol (NaOH);
-2 mol            -2 mol hence, X mol (NaOH) = 4 * 2/2 = 4 mol.
Let’s calculate the mass of sodium hydroxide:
m (NaOH) = Y * M = 4 * 39.9 = 159.6 g.
Answer: the mass of sodium hydroxide is 159.6 g.