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.

Md Ariful Islam

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Md Ariful Islam

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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

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)

$(65xx10^-3g) /(79.9040000005 ) = 0.00081347617"mol"$

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$

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