How many grams of cah2 are needed to generate 143 l of h2 gas if the pressure of h2 is 827 torr at 22 ∘c? g. com?

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