c6h5brl

  1. J

    C6H6(l) Br2(l) = C6H5Br(l) + HBr(g)

    C6H6(l) Br2(l) = C6H5Br(l) + HBr(g) My question is if 18.3 grams of Benzene (C6H6) reacts with 3.21 grams of Bromine (Br2) how many grams of Bromobenzene (C6H5Br) will be produced? I have tried this multiple ways and I am really lost when it comes to setting this up. I tried converting the...