The mass of water formed is
<u><em>calculation</em></u>
Use the ideal gas equation to calculate the moles of NH3 and O2
that is Pv= n RT
where; P= pressure,
V= volume,
n = number of moles,
R=gas constant = 0.0821 l .atm/ mol.K
make n the formula of the subject by diving both side by RT
n = PV /RT
The moles of NH3
n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles
The moles of O2
=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles
write the reaction between NH3 and O2
4 NH3 + 5 O2 →4 No +6H2O
from equation above 0.766 moles of NH3 reacted to produce
0.766 x 6/4 =1.149 moles of H2O
0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20
since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles
mass of H2O = moles x molar mass
from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol
mass = 18 g/mol x 0.9408 moles= 16.93 grams
