Sea water contains `65 xx10^-3` `"g/L"` of bromide ions. If all the bromide ions are converted to produce `Br_2`,how much sea water is needed to prepare 1 kg of `Br_2`? please tell the method.

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`