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.

Question

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.

Natalie Anton · Accepted Answer

#15384.6156003L# of sea water

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#

Jose Ayala · Answer

undefined To find the amount of sea water needed to produce 1 kg of Br2, first calculate the moles of Br2 needed using the molar mass of Br2. Then, use the stoichiometry of the reaction between Br- ions and Br2 to find the moles of sea water required. Finally, convert the moles of sea water to liters using the density of sea water.