Need help with C
- The following chemical equation represents the reaction between benzene (C6H6) with bromine.
C6H6(l) +Br2(l) à C6H5Br(l) + HBr(g)
If 65.0 g of C6H6 is mixed with 150.0 g of Br2
a.) What is the limiting reactant? (C =12.01 g/mol, H = 1.008 g/mol, Br = 79.90 g/mol)
Mass of C6H6 = 65.0 g
Molar mass of C6H6
C= 6 x 12.011 = 72.066
H = 6 x 1.008 = 6.048
72.066 + 6.048 = 78.114 g/mol
Mass of Br2 =150.0 g
Molar mass of Br2
Br = 2 x 79.904 = 159.808 g/mol
Moles of C6H6 = 0.8321 mol C6H6
Moles of Br2 = 0.9386 mol Br2
Stoichiometric ratio 1:1
0.8321 mol C6H6 =0.832
0.9386 mol Br2 = 0.939
The limiting reactant is C6H6
b. How many grams of C6H5Br is formed in the reaction (theoretical yield)?
Because Stoichiometric ratio between C6H6 and C6H5Br is 1:1
Then 0.832 mol C6H6 = 0.832 C6H5Br
Molar mass of C6H5Br
C = 6 x 12.011 = 72.066
H = 5 x 1.008 = 5.04
Br = 1 x 79.904 = 79.904
Molar mass = 72.066+5.04+79.904= 157.01
0.832 mol g x 157.01= 130.64g = 131 g of C6H5Br is formed
c. Find the actual mass of C6H5Br that is formed, if the percent yield is 88.5%?
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