[portugal]

1. When KBr reacts with conc. H2SO4 some bromine and SO2 are evident as well as the main product (HBr). The number of moles of KBr which reacts with one mole of conc. H2SO4 in this side reaction is:

 a. 1.33 mol

 b. 0.75 mol

 c. 1.00 mol

 d. 2.00 mol

 e. 0.50 mol

[Borek]

Please read forum rules. Start with reaction equation.

[portugal]

thats exactly what i am having problems with establishing the equation and once someone can help me out with that i can do the rest

[Borek]

You know reactants, you know products. Write skeletal reaction, then it is just a matter of balancing.

[portugal]

so would you say this is the correct answer then:

2 KBr + 2 H2SO4 = 4 HBr + -0 SO2 + -1 Br2 + 2 KSO4

Therefore the answer would be 1?

[Borek]

1. No such thing as KSO₄

2. You are looking for side reaction - so your skeletal should not contain main product (HBr)

3. You have two weird coefficients - one is negative (all should be positive) one is zero (which means there is no SO₂ in products, yet you were specifically told it IS present, so your equation doesn't describe the reaction in question).

What you were told is that side reaction is between KBr + H₂SO₄ and it yields Br₂ + SO₂:

KBr + H₂SO₄ -> Br₂ + SO₂

However, this can't be right for an obvious reason - no K and H on the right side. The most logical guess is that K becomes K₂SO₄. What about hydrogen? Hmm, that's more complicated, however, you have hydrogen and at the same time SO₄^2- gets converted to SO₂ - so it looses oxygen. Hydrogen and oxygen means water, so your skeletal reaction should look like:

KBr + H₂SO₄ -> Br₂ + SO₂ + K₂SO₄ + H₂O

And you should balance it now.