How would i determine the products and balance the following reactions , plus how would i classify them?

Question

1. KOH(aq) + FeCl3(aq) ----
2. Al(s) + Br2(l) -----
3.C5H10(l) (pentane) + O2(g) -----

Answer 1

You determine the products by looking at the reactants.

In num. one there are two compounds so they will "switch"
K is a metal and therefore will be drawn to the nonmetal of the other compound. what is left goes together.
To balance this, just make sure you have as many numbers of each element on both sides of the equation.

In num. two there are two elements. Combine them. Al is in 3A so it has an oxidation value of +3. Br is in 7A therefore it has an oxidation value of -1 (it must gain one electron) Therefore you will need three Br for one Al. You end up with AlBr3
You balance it the same as #1

For #3 you have C, H2, and O2 therefore you have a combustion. (the products will be CO2 + H2O)
Balance the same as the others

Answer 2

1. 3KOH(aq) + FeCl3(aq) ===> Fe(OH)3(s) + 3 KCl(aq) Double displacement
2. 2Al(s) + 3Br2(l) ===> 2AlBr3(s) Combination
3. 2C5H10(l) + 15O2 ===> 10CO2 + 10H2O Oxidation reduction

(C5H10 would be pentene or cyclopentane, not pentane.)