Answer: 26.54 grams
Explanation:
To calculate the moles :
is the limiting reagent as it limits the formation of product and
is the excess reagent
According to stoichiometry :
As 1 moles of give = 3 moles of
Thus 0.369 moles of give =
of
Mass of
Thus 26.54 g of will be produced from the given mass.
Answer:
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:
With further simplification, we have:
Then, we have:
Therefore,
Using the expansion:
Therefore,
Thus:
equation (1)
Using the virial equation of state:
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2