How many grams of water and sodium do you need to take to prepare 120g of a 30% solution?

2Na + 2H2O = 2NaOH + H2

Find the mass and amount of the sodium hydroxide substance:

m (NaOH) = w (NaOH) * m (solution) = 0.3 * 120 = 36 g;

n (NaOH) = n (NaOH) / M (NaOH) = 36/40 = 0.9 mol;

According to the stoichiometry of the reaction, 0.9 mol of water is required to obtain sodium hydroxide. The mass of the reacted water will be:

mreacted (H2O) = n (H2O) * M (H2O) = 0.9 * 18 = 16.2 g:

The total mass of water will be:

m (H2O) = m (solution) – m (NaOH) + m reacted (H2O) = 120 – 36 + 16.2 = 100.2 g;

Find the mass of sodium:

n (Na) = n (NaOH) = 0.9 mol;

m (Na) = n (Na) * M (Na) = 0.9 * 23 = 20.7 g.

Answer: m (H2O) = 100.2 g; m (Na) = 20.7 g.