A 2.88 g mixture of calcium metal and calcium hydride reacts completely with excess water and liberates 2.464 L of hydrogen at STP. Using Ca = 40, H = 1 and molar volume 22.4 L/mol, calculate the mass percentage of $CaH X_{2}$ in the mixture, to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

Answer: 58 %

Let

x

mol Ca and

y

mol

CaH X_{2}

Ca + 2 H X_{2} O Ca (OH) X_{2} + H X_{2}

gives 1 mol

H X_{2}

per Ca and

CaH X_{2} + 2 H X_{2} O Ca (OH) X_{2} + 2 H X_{2}

gives 2 mol per

CaH X_{2}

. Then

40 x + 42 y = 2.88

and

x + 2 y = 2.464/22.4 = 0.1100

. Solving gives

x = 0.0300

y = 0.0400

, so the mass of

CaH X_{2}

1.68 g

and the percentage is

1.68/2.88 \times 100 = 58.33 \approx 58%