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How much Aluminum Chloride in grams is needed to completely precipitate 1.74 grams of K2SO4 (FW=174 g/mol)?

2 Answers
Mar 5, 2018

Answer:

The question is unsound...

Explanation:

We could change it to how much #"barium nitrate"# is required to react with #1.74*g# of #"potassium sulfate..??"#

And here, we would write the stoichiometric equation...

#Ba(NO_3)_2(aq)+K_2SO_4(aq)rarrBaSO_4(s)darr+2KNO_3(aq)#

And thus moles of potassium sulfate is equal to moles of barium nitrate.

#"Moles of potassium sulfate"=(1.74*g)/(174.26 *g*mol^-1)=0.010*mol#

And thus we require an equivalent quantity of barium nitrate...

#=0.010*molxx261.34*g*mol^-1=2.61*g#...

Mar 5, 2018

Answer:

Here it is: 0.89 grams

Explanation:

#3K_2SO_4 + 2AlCl_3 -> Al_2(SO_4)_3 + 6KCl#

is the balanced chemical reaction.

Now you have #1.74/174 = 0.01# moles of #K_2SO_4#

Therefore you need 0.0067 mole of #AlCl_3#

(MW of it is 133.5 grams). Therefore total amount needed to complete this reaction is #=0.0067times133.5 = 0.89# grams.

You need 0.89 grams of #AlCl_3# to precipitate 1.74 grams of #K_2SO_4#