Question

Both solid LiH and CaH2 react with water to produce hydrogen gas and the corresponding hydroxide, LiOH or Ca(OH)2. A 0.850g sample of a mixture of LiH an CaH2 produces 1.200 liters of H2 at STP. What percentage of the starting mixture was LiH? (Give both mole percent and percent by weight)

Answer 1

The Reactions can be represented by

LiH (s) + H2O (l) --> LiOH (aq) + H2 (g) (1)

CaH2 (s) + 2H2O (l) --> Ca(OH)2 (aq) + 2H2 (g) (2)

Let the mass of LIH = x, mole of LIH= mass/ molecularr mass = x/7.95 moles= 0.126x moles

From reaction (1) 1 mole of LIH produces =1 mole of H2,

Hydrogen formed = 0.126x moles

Then the mass of Ca(OH)2= 0.85-x,, moles of Ca(OH)2= (0.85-x)/74 moles ,

From stoichiometry Eq(2), moles of Hydrogen produced = 2*(0.85-x)/74= 0.027* (0.85-x )

Total moles of hydrogen produced = 0.126x+ 0.027*(0.85-x)=0.126x+ 0.02295-0.027x =0.099x+0.02295 (3)

1 mole of any gas at STP produces 22.4 liters

1.2 liters corresponds to 1.2/22.4 moles =0.053571 moles

Equating this to Eq.3

0.053571 =0.099x +0.02295

0.053571- 0,02295 =0.099x

0.030261=0.099x

x= 0.31 gms

Hence mass off LiH =0.31 gms,mass % of LiH =(0.31/0.85)*100= 36.4%

moles of LIH = 0.31/7.95= 0.038994

moles of CaH2 =(0.85-0.31)/74 =0.007297

Total moles = 0.038994 +0.007297 =0.046291

mole % of LIH = (0.038994/0.046291)*100=84.2%