Part A itm18 , Chlorine gas reacts with fluorine gas to form chlorine trifluorid

Question

Part A itm18 , Chlorine gas reacts with fluorine gas to form chlorine trifluoride.
Cl2(g)+3F2(g)2ClF3(g)
A 2.10 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 855 mmHg .

Identify the limiting reactant and determine the theoretical yield of ClF3 in grams.

Part B , itm15 . The metabolic oxidation of glucose, C6H12O6, by oxygen gas in our bodies produces water and carbon dioxide, which is expelled from our lungs as a gas. Calculate the volume of dry carbon dioxide produced at body temperature (37 C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.

Part C , item 14 . Calculate the volume of dry carbon dioxide produced at body temperature (37 C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.

a)What volume of H2 gas (in L), measured at 748 mmHg and 86 C, is required to synthesize 24.6 g CH3OH?

b)How many liters of CO gas, measured under the same conditions, is required?

Explanation / Answer

Given: Cl2 + 3 F2 2 ClF3

Pressure of Cl2 =337mg

Pressure of F2 = 855mg

So, 1 mole of Cl2 react with 3 moles of F2.

So, 337mg of Cl2 shall react will F2 = 337 x 3 = 1011mg.

But present F2 = 855mg. So, F2 is the limiting reagent.

Now, 3 moles F2 produce ClF3 = 2moles

855mmHg F2 produce ClF3 = 2/3 x 855 = 570mm Hg

Pressure of ClF3 formed = 570 mmHg

Now, Use PV = nRT

570 mmHg F2 x 2.1L = n x 62.36 LmmHg/Kmol x 298 K

n = 0.0644 mol ClF3

Mass of ClF3 obtained = no. of moles x molar mass = 0.0644 x 92.4482 g/mol) = 5.955 g ClF3 = Theoretical Yield.

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