When 7.6 g of a mixture of potassium and sodium hydrides were treated with excess water, 5.6 dm3 of hydrogen was released.

When 7.6 g of a mixture of potassium and sodium hydrides were treated with excess water, 5.6 dm3 of hydrogen was released. Determine the mass fraction of hydrides in the mixture.

And so, first we write down the equations of reactions that pass by the condition:
KH + H2O = KOH + H2
NaH + H2O = NaOH + H2
Since hydrogen is a gas, we will use the formula
n (H2) = V (H2) / Vm
n (H2) = 5.6 / 22.4 = 0.25 mol
Let be
x – n mol KH
y – n mol NaH
Then we compose an equation, or rather a system of equations, based on the initial conditions
40x + 24y = 7.6
x + y = 0.25
Having solved the system, we get
x = 0.1 mol
y = 0.
Find the mass
m (KH) = n (KH) * M (KH)
m (KH) = 0.1 * 40 = 4 g
m (NaH) = 0.15 * 24 = 3.6 g
Let’s find the mass fractions by the formulas
w = m (in-va) / m (solution) * 100%
w (KH) = 4 * 100 / 7.6 = 52.6%
w (NaH) = 100 – 52.6 = 47.4%
Answer: 52.6%, 47.4%