We know that the sea water contains #(65xx10^-3"g of bromide ions")/L #
If you want #Br_2# the following reaction takes place
#Br^-) + Br^-) = Br_2 + 2e^-#
Well this reaction cannot take place but this can be a half reaction
But this reaction can take place
#2"HBr" +"H"_2"SO"_4 rightleftharpoons "Br"_2 + "SO"_2 + 2"H"_2"O"#
So this like stoichiometry
1mol of Br- reacts with 1mol of Br- to form 1 mol of #Br_2#
or 2mol of #"Br"^-# react to form 1 mol of #Br_2#
First calculate the amount of moles of #"Br"_2# formed when #65xx10^-3"g of bromide ions"# react
Recall
#"grams of substance"/"molar mass of substance" = "moles""
"(1)#
#(65xx10^-3g) /("molar mass of Br"^-)#
Since the Br- has an extra electron
#"Molar mass of Br"^-) = "Molar mass of Br" + "Molar mass of electron"#
#(79.904g)/(mol) + (5.485 799 090 70 * 10^-10g)/"mol" = "79.9040000005g/mol"#
Now plug the variables equation (1)
#(65xx10^-3g) /(79.9040000005 ) = 0.00081347617"mol"#
The mole ratio of bromine ions reacted to #"Br"_2# formed is
#1mol : 0.5mol#
Therefore solving for ratio
#0.00081347617 : x = 1 : 0.5#
#0.00081347617 : 1/2*0.00081347617 = 1 : 0.5#
0.00040673808 mol of #"Br"_2# is formed
We need 1kg of #"Br"_2# so we have to find the no. of moles in 1kg of #"Br"_2#
Rearranging equation 1 we get
#"grams of substance" = "moles" xx "molar mass"#
#1000g = "molar mass of Br" xx 2 xx x#
# 79.904*2x = 1000#
#159.808x = 1000#
#"moles" = 1000/159.808#
#= "6.25750901081moles"#
If #("0.00040673808 mol of Br"_2)/("1L of sea water")#
Then
#("6.25750901081moles of Br"_2)/("x") = ("0.00040673808 mol of Br"_2)/("1L of sea water")#
#x# is the desired amount of sea water
#6.25750901081 = 0.00040673808x#
#x = 6.25750901081/0.00040673808#
# 15384.6156003L#
