Okay
Mr (H2O)= 18g
therefore moles of H2O
is 720.8/18= 40.04mol
the ratio of H2 to O2 to H2O is
2 : 1 : 2
so moles of H2 is same as H2O here
H2= 40.04moles
moles of O2 is half
so 40.04 x 0.5
20.02moles
grams of O2 is
its moles into Mr of O2
that's 20.02 x 32 = 640.64g
The upper surface of the zone of saturation is called the water table
The balanced chemical
reaction will be:
C4H8 + 6 O2 --> 4 CO2 + 4 H2O
We are given the amount of butene being combusted. This will be our
starting point.
136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
