a.What is the theoretical yield of bromobenzene in this reaction as soon as 30.0g the benzene reacts with 65.0 g that bromine?
ok this trouble is a tiny tricky...
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1. V these varieties of problems constantly ALWAYS balance the equation first, it will really help you later in the problem.
The well balanced equation is
1 C6H6 + 1 Br2 --> 1 C5H6Br + 1 HBr
2. Ok, next you must recognize which reactant is in excess and also which is limiting...you cannot always determine this by the grams of each reactant. Girlfriend MUST transform grams to moles.
30.0 g C6H6 (1 mol C6H6 / 78.54 g C6H6) = 0.382 mol C6H6
* 78.54 is the atomic mass of C6H6
60.0 g (1 mol Br2 / 159.8 g Br2) = 0.375 mol Br2
* 159.8 is the atomic mass that C6H6
3. Because there are much less moles the Br2, than there are C6H6, Bromine is the limiting reactant and also will be offered to calculation the yield.
To calculate yield collection up this equation:
60.0 g Br2 (1 mol Br2 / 159.8 g Br2)(1 mol C6H5Br / 1mol Br2)(157 g C6H5Br / 1 mol C6H5Br) = 58.95 g C6H5Br
* because that the 1 mol C6H5Br / 1 mol Br2 part, ns just got those number from the balanced equation. Ns looked at the coefficients (numbers prior to each compound). Because that every one mol of Br2, one mole of C6H5Br is produced.
Now 58.95 g is the theoretical yield!
4. To uncover the percent yield:
(actual yield / theoretical yield) x 100%
So... (56.7/58.95) x 100% = 96.18%
I expect this helped! great luck!
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WHEN benzene (C6H6) reacts with bromine (Br2) bromobenzene(C6H5Br) ...?
WHEN benzene (C6H6) reacts v bromine (Br2) bromobenzene(C6H5Br) is obtained:
C6H6 + Br2 → C6H5Br + HBr
a.What is the theoretical yield of bromobenzene in this reaction when 30.0g that benzene reacts with 65.0 g that bromine?
b.B. If the actual productivity of bromobenzene...
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