Question

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

Answer 1

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`...

Answer 2

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`