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;