# With the action of excess water on 7.6 grams of a mixture of sodium hydride and potassium hydride

**With the action of excess water on 7.6 grams of a mixture of sodium hydride and potassium hydride, 5.6 liters of hydrogen were obtained. Determine the mass composition of the initial mixture.**

NaH + H2O = NaOH + H2

KH + H2O = KOH + H2

Let the mass of sodium hydride be x, the mass of potassium hydride – y.

By condition: x + y = 7.6 g

Let’s find the molar mass of hydrides.

n (NaH) = m (NaH) / M (NaH) = x / 24

n (KH) = m (KH) / M (KH) = y / 40

n (NaH) = n1 (H2) – in the first reaction.

n (KH) = n2 (H2) – in the second reaction.

Let’s find the total amount of moles of released hydrogen.

n (H2) = V (H2) / Vm = 5.6 / 22.4 = 0.25 mol

n1 (H2) + n2 (H2) = n (H2) = 0.25

x / 24 + y / 40 = 0.25

We solve the system of equations:

x + y = 7.6

x / 24 + y / 40 = 0.25

Hence y = 3.86 g, x = 3.74 g.