Answer to Question #86441 in General Chemistry for Ainslea

Question

Benzene reacts with bromine to form bromobenzene, C6H5Br.
C6H6(L) + Br2(L) ---> C6H5Br(L) + HBr (g)

A) What is the maximum amount of C6H5Br that can be formed from the reaction of 7.50g of C6H6 with excess Br2?

B) A competing reaction is the formation of dibromobenzene, C6H4Br.
C6H6(l) + 2Br2(L) ----> C6H4Br2(L) + 2HBr(G)
If 1.25 g of C6H4Br2 was formed by the competing reaction, how much C6H6 was not coverted to C6H5Br?

Answer 1

a) C₆H₆(l) + Br₂(l) -> C₆H₅Br(l) + HBr(g)

n= m/M

n(C₆H₆) = 7.50 g / 78.11 g/mol = 0.096 mol

according to equation mole ratio n(C₆H₆):n(C₆H₅Br) = 1:1, then n(C₆H₅Br) = n(C₆H₆) = 0.096 mol

m(C₆H₆Br) = 0.096 mol * 157.01 g/mol =15.07 g

b) C₆H₆(l) + 2Br₂(l) -> C₆H₄Br₂(l) + 2HBr(g)

n= m/M

n(C₆H₄Br₂) = 1.25 g/ 235.9 g/mol = 0.0053 mol

According to equation mole ratio n(C₆H₆):n(C₆H₄Br₂) = 1:1, then n(C₆H₆) = n(C₆H₄Br₂) = 0.0053 mol

m(C₆H₆) = 0.0053 mol * 78.11 g/mol = 0.412 g

Answer: a) 15.07 g; b) 0.412 g