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.

1 Answer
May 26, 2017

Answer:

#15384.6156003L# of sea water

Explanation:

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#