# 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. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.