Question

a sample of cesium metal reacted completely with water, evolving 48.1 ml of dry H2 at 19C and 768 mmHg. what is the equation for the reaction? what was the mass of cesium in the sample ?

Answer 1

The reaction  for  equation is    written  as  follows

2Cs (s) +  2H2O(l) =  2CsOH (aq) +  H2  (g)

The  mass  of  cesium  is  calculated  as  follows

Find  the  moles  of  H2  by use  of  ideal  gas  equation  that is  Pv = nRT
where  n  is  the  number of   moles

 n  =  PV/RT
P=  48.1  ml    in  liters  =  48.1  /1000= 0.0481 l
T=  19 +  273.15 =  292.15K
P=  768mm hg
R (gas  Constant)=  62.364 l.mmhg/k.mol

n= ( 768mmhg  x0.0481 L) /( 62.364 L.mm hg/k.mol  x  292.15k) =  2.028  x10^-3  moles

by   use  of  mole  ratio from  reacting  equation between  Cs to  H2    which  is   2 :1  the  moles  of Cs  is therefore  =  ( 2.028  x10^-3)  x  2 =  4.05  x10^-3 moles

mass  of  cs is therefore  =  moles  x  molar  mass  of  cs(  132.9g/mol)

=( 4.05  x10^-3)mol  x  132.9 g/mol  =  0.539  grams