For the reaction kclo2⟶kcl+o2 kclo2⟶kcl+o2 assign oxidation numbers to each element on each side of the equation. k in kclo2: k in kclo2: k in kcl: k in kcl: cl in kclo2: cl in kclo2: cl in kcl: cl in kcl: o in kclo2: o in kclo2: o in o2: o in o2: which element is oxidized? kk oo clcl which element is reduced? oo kk cl

K in KClO2 = +1

Cl in KClO2 = +3

O in KClO2 = -2

K in KCl = +1

Cl in KCl = -1

O in O2 = 0

Chlorine is going from +3 to -1 so it is being reduced

Oxygen is going from -2 to 0 so it is being oxidized

Explanation:

Potassium is a constant +1

Chlorine could be -1, +1, +3, +5, or +7

Oxygen can be either -2 or +2

Every reactant has to equal zero if there is not a given charge.

Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g