Determine mass of water formed when 12.5 l nh3(at298k and 1.50atm) is reacted with 18.9l of o2 (at 323k and 1.1atm)

The  mass  of water formed  is

calculation

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

idk

explanation: