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Chlorine gas and fluorine gas react to form chlorine trifluoride gas. Initially, 1.75...
- Chlorine gas and fluorine gas react to form chlorine trifluoride gas. Initially, 1.75 mol of chlorine and 3.68 mol of fluorine are combined. After the reaction is complete, 192 g of chlorine trifluoride are obtained. What is the percent yield of the reaction? Hint: you will need the balanced chemical equation for this reaction.
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The balanced chemical equation for the reaction is:
Cl2 + 3F2 = 2ClF3; relevant stoichiometric relationships are:
1 mole, 3 moles, 2 moles = 2×92.5 = 185 g
1.75 moles of chlorine reacts with 3×1.75=5.25 moles of fluorine, but in actual practice 3.68 moles of fluorine is present. Clearly fluorine is the limiting reagent here.
Again, 3 moles of fluorine reacts to produce 185g of ClF3.
So, 3.68 moles of fluorine should ideally produce 185×3.68/3 = 226.93g ClF3.
The actual yield is 192g, hence percent yield of the reaction is 192×100/226.93 = 84.6%
Posted by llltkl on April 1, 2013 at 3:56 PM (Answer #1)
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