# How much sodium oxide needs to be dissolved in water to get 8 g of NaOH?

**How much sodium oxide needs to be dissolved in water to get 8 g of NaOH? How many times will the mass of sodium oxide be greater than the mass of water?**

The interaction of sodium oxide with water occurs in accordance with the following chemical reaction equation:

Na2O + H2O = 2NaOH;

1 mole of oxide reacts with 1 mole of water. As a result, 2 moles of alkali are formed.

Let’s calculate the chemical amount of a substance that is contained in 8 grams of alkali.

M NaOH = 23 + 16 + 1 = 40 grams / mol;

N NaOH = 8/40 = 0.2 mol;

To prepare such an amount of alkali, you need to take 0.1 mol of sodium oxide and 0.1 mol of water.

Let’s calculate the masses of these substances.

M Na2O = 23 x 2 + 16 = 62 grams / mol;

m Na2O = 0.1 x 62 = 6.2 grams;

M H2O = 2 + 16 = 18 grams / mol;

m H2O = 0.1 x 18 = 1.8 grams;

Let’s calculate how many times the mass of sodium oxide will be greater than the mass of water.

K = m Na2O / m H2O = 6.2 / 1.8 = 3.44 times;