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uysha [10]
3 years ago
13

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal?

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
8 0
The balanced chemical reaction is written as:

<span>2K + F2 ---> 2 KF

We are given the amount of potassium metal to be used in the reaction. This will be the starting point for the calculation. We do as follows:

23.5 g K ( 1 mol / 39.1 g ) ( 1 mol F2/2 mol K ) ( 22.4 L / 1 mol ) = 13.46 L F2 

</span>
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The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:
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To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
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*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
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Second, you can change the moles of C2H5OH to moles of C6H12O6..
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Third, change moles of C6H12O6 to grams...
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2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!
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