You are given 48g of water how many moles do you have?
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.
<u>Answer:</u> The mass of produced is 2.52 g
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
......(1)
- <u>For hexane:</u>
Given mass of hexane = 1.72 g
Molar mass of hexane = 86.18 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
Given mass of oxygen gas = 8.0 g
Molar mass of oxygen gas= 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of hexane follows:
By stoichiometry of the reaction:
If 2 moles of hexane reacts with 19 moles of oxygen gas
So, 0.020 moles of hexane will react with = of oxygen gas
As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hexane is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of hexane produces 14 moles of
So, 0.020 moles of hexane will produce = of
We know, molar mass of = 18 g/mol
Putting values in above equation, we get:
Hence, the mass of produced is 2.52 g