Science, asked by manoranjan11, 1 year ago

21.
The percent yield for the following reaction carried out in carbon tetrachloride (CCI) solution is 80%
Br, +CI- 2BCI
(a) What amount of BrCl would be formed from the reaction of 0.025 mol Br, and 0.025 mol CI?
(b) What amoumt of Br, is left unchanged?
LA​

Answers

Answered by gogiya167
13

Answer:

Explanation: If the yield were 100%, the mass of the one product would be the same as the combined mass of the reactants, right? (Conservation of mass principle)  

Figure out how many grams 0.025 mol Br2 is and how many grams 0.025 mol Cl2 is and add them up. That would have been 100% yield.  

Br2 = (0.025)(2)(79.909) = 4gm.  

Cl2 = (0.025)(2)(35.453) = 1.773gm.  

Total = 5.773gm This would have been 100% yield.  

You got 80% of this or (.80)(5.773) = 4.62grams product

For part b)  

For every 2 moles of BrCl you made, you had to use 1 mole of Br2--just half as much. [Information obtained from coefficients.]  

Find how many moles of BrCl you made.  

A mole of BrCl is 79.909g + 35.453g = 115.362 grams  

You have 4.62g so: 4.62/115.362 = 0.04 moles of BrCl that you made.  

You must have used up just half that many moles of Br2. [From our reasoning about the coefficients.]  

So 0.02 divided by 2 = 0.02 moles Br used up  

Therefore you have 0.025-0.02 = 0.005 moles of Br2 left.

Weight of Br2 = 2*0.005*79.909=0.7999 gm left unchanged.

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