# A mixture of sodium oxide and potassium oxide was dissolved in water, where the mass fraction of sodium is 46%

**A mixture of sodium oxide and potassium oxide was dissolved in water, where the mass fraction of sodium is 46%. Determine the mass of the initial mixture if 175 ml of a 2 molar solution of sulfuric acid were consumed to completely neutralize the resulting mixture**

Decision:

M (Na2O) = (2 * 23) + 16 = 62 g / mol;

M (K2O) = (2 * 39) + 16 = 94 g / mol.

From the formula of sodium oxide Na2O, it follows that 46 parts by weight of Na account for 16 parts by weight of O, i.e. 46% Na in the mixture of oxides will account for 16% O. Then the mixture of sodium and potassium oxides will contain 62% (46 + 16 = 62) Na2O. Accordingly, the K2O content in the mixture will be 38% (100 – 61 = 100).

The reaction equation has the form:

Me2O + H2SO4 = Me2SO4 + H2O

It follows from the reaction equation that 1 mole of sulfuric acid is consumed in the neutralization of 1 mole of sodium or potassium oxides.

We calculate the amount of sulfuric acid according to the condition of the problem, we get:

(175ml * 2mol / 1000ml) = 0.35 mol. Therefore, n (mixture) = 0.35 mol.

The mixture contains 0.35 mol of sodium and potassium oxides.

We denote the mass of the oxide mixture by xg.

Then

n (Na2O) = (62% * xg) / (100% * 62g / mol) = 0.01x mol;

n (K2O) = (38% * xg) / (100% * 94g / mol) = 0.0004x mol.

Now we calculate the mass of the mixture of sodium and potassium oxides using the algebraic equation, we get:

0.01x + 0.004x = 0.35;

0.014x = 0.35;

x = 0.35 / 0.014 = 25g; m ((mixture) = 25g.

From here

m (Na2O) = [m ((mixture) * w% (Na2O)] / 100% = [(25g * 62%)] / 100% = 15.5g;

m (K2O) = [m ((mixture) * w% (K2O)] / 100% = [(25g * 38%)] / 100% = 9.5g.

Answer: m (mixture) = 25g.