An 0.850 g sample of a mixture of LiH and CaH2 was reacted with H2O and the H2 gas produced, 1.20 L, was collected at STP. What mass % of the original mixture was LiH?

Answer 1

work it in two parts. let V be the volume of H2 produced in the CaH2 reaction, and 1.2-V be the volume in the LiH reaction. Let m be mass in the CaH2 reaction, and .850g -m be the mass in the LiH reaction.
You will get two equations, one unknown out of this.
CaH2 reaction:
CaH2 + 2H2O = Ca(OH)2 + 2H2
H2: P:1 atm T: 273K V: V
PV=nRT
nCa=PV/RT
n= (1*V)/(0.08206*273)
n= you do it.
LiH reaction:
same method, solve for NLi
you will get N in terms of V

now convert each of the N's to grams (by multiplying each by formaula masses of the hydrides), and set each to a funcion of m.
you should have time then to find the ratios of the two masses, and solve for percent. A bit of algebra is required.

Answer 2

LiH + H2O ==> LiOH + H2
CaH2 + 2H2O ==> Ca(OH)2 + 2H2
You must write two equations and solve them simultaneously. NOTE; ALL OF THESE NUMBERS ARE ESTIMATES; YOU SHOULD RECALCULATE ALL OF THEM.

Let Y stand for grams LiH.
and Z stand for grams CaH2.
Then equation 1 is Y + Z + 0.850

How many mols is 1.50 L @ STP. That's 1.50/22.4 = aprox 0.067
For the second equation you want to write an equation that equates mols H2 from LiH and mols H2 from CaH2 = 0.067

Below I will use mm for molar mass.

So mols H2 from LiH will be
[(Y/mm LiH)] x [(1 mol H2/1 mol LiH)]

and mol H2 from CaH2 will be
[(Z/mm CaH2)] x [2 mols H2/1 mol CaH2)]

Solve for Y = grams LiH.

Then % LiH = (g LiH/mas sample)*100
Post your work if you get stuck.