Consider the reaction between sodium metal and fluorine gas to form sodium fluoride. using oxidation states, how many electrons would each sodium atom lose, and how many electrons would each fluorine atom gain? #e- each sodium atom would loose? #e- each fluorine atom would gain? how many sodium atoms are needed to react with one fluorine molecule?

Explanation:

Sodium is the element of the group 1 and period 3 which means that the valence electronic configuration is .

Sodium is most likely to loose 1 electron so as to gain noble gas configuration (Stable oxidation state = +1) .

Fluorine is the element of the group 17 and period 2 which means that the valence electronic configuration is .

Fluorine is most likely to gain 1 electron so as to gain noble gas configuration (Stable oxidation state = -1) .

Thus, 1 atom of sodium losses 1 electron to 1 atom of fluorine and the atoms of  fluorine accepts this one electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Thus, the formula of sodium fluorine is .

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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