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
Considering the reaction stoichiometry and ideal gas law, the mass of water formed is 16.956 grams.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P×V = n×R×T
In this case, you know:
Then:
1.50 atm× 12.5 L= n× 0.082 [tex]\frac{atmL}{molK}[/tex]×298 K
Solving:
[tex]n=\frac{1.50 atmx 12.5 L}{0.082 \frac{atmL}{molK}x298 K}[/tex]
n= 0.767 moles
Then:
1.10 atm× 18.9 L= n× 0.082 [tex]\frac{atmL}{molK}[/tex]×323 K
Solving:
[tex]n=\frac{1.1 atmx 18.9L}{0.082 \frac{atmL}{molK}x323 K}[/tex]
n= 0.785 moles
On the other side, the balanced reaction between NH₃ and O₂ is:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:
The limiting reagent will be the one that is exhausted first in the reaction, so once it is finished, the reaction ends. So the limiting reagent is one that limits the reaction and the amount of shaped product.
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 3 moles mole of NH₃ reacts with 5 moles of O₂, 0.767 moles of NH₃ react with how much moles of O₂?
[tex]moles of O_{2} =\frac{0.767 moles of NH_{3} x 5moles of O_{2} }{3moles of NH_{3} }[/tex]
moles of O₂= 0.96
But 0.96 moles of O₂ are not available, 0.785 moles are available. Since you have less moles than you need to react with 0.767 moles of NH₃, oxygen O₂ will be the limiting reagent.
Then, it is possible to determine the amount of moles of H₂O produced by another rule of three: if by stoichiometry 5 moles of O₂ produce 6 moles of H₂o, if 0.785 moles of O₂ react how many moles of H₂O will be formed?
[tex]moles of H_{2}O =\frac{0.785 moles of O_{2} x 6moles of H_{2} O}{5moles of O_{2} }[/tex]
moles of H₂O=0.942
The molar mass of H₂O being 18 g/mole, the produced mass of the compound is calculated as:
[tex]0.942 molesx\frac{18 grams}{1 mole} =[/tex] 16.956 grams
Finally, the mass of water formed is 16.956 grams.
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