The grams of caH2 that are needed to generate 143 L of H2 is calculated as follows
by use of ideal gas equation Pv=nRT where n is number of moles, calculate the moles of H2 produced
by making n the subject of the formula
n= PV/RT
p= 827 torr
R(gas constant)= 62.36 L.torr/mol.K
T= 273 +22=295 k
V=143 L
n =(827 torr x143 L)/ 62.36 L.torr/mol.k x295 k) =6.43 moles
write the reacting equation
that is caH2 +2H2O= Ca(OH)2 + 2H2
by use of mole ratio between CaH2 to H2 which is 1 :2 the moles of CaH2 = 6.43 x1/2=3.215 moles
mass of CaH2 is therefore= moles of CaH2 x molar mass of CaH2
=3.215 moles x 42 g/mol = 135.03 grams
