Give an example of how you could test Charles Law?
1 answer:
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Answer:
2.5 % butane, 42.2 % pentane and 55.3 % hexane
Explanation:
Hello,
In this case, the mass balance for each substance is given by:
Whereas accounts for the fractions at the outlet distillate and
for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:
So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:
The total distillate flow:
And the total bottoms flow:
Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:
Then, the molar fraction of pentane and hexane:
Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.
NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.
Regards.
a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :
volume NO at 1273 K and 1 atm
b. 15 L NH3 at STP ( 1mol = 22.4 L)
mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :
mass H2O(MW = 18 g/mol) :
c. mol NO at 1273 K and 1 atm :
mol ratio of NO : O2 = 4 : 5, so mol O2 :
Volume O2 at STP :