Question

11.2dm3 of ethene at STP has moles of H2 atoms

Answer 1

Answer:

Explanation:

V   =    11.2dm^3                 1dm^3 = 1L

     =   11.2l

at stp, P = 1atm, T =273k

ideal gas eQ

PV =  nRT

n  = PV/RT

  =  (1* 11.2(L))/0.083 *273  =  0.49 mol of ethene(c2h4)

1mol of c2h4 contain 4 moles of h atoms

0.49 mol of c2h4 contain = 4*0.49 = 1.97mol  Ans