Consider this mechanism: step 1: clo-(aq) + h2o(l) hclo(aq) + oh-(aq) (fast) step 2: i-(aq) + hclo(aq) hio(aq) + cl-(aq) (slow) step 3: oh-(aq) + hio(aq) h2o(l) + io-(aq) (fast) what is the overall reaction?

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

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