What mass of precipitate should be obtained by the reaction between masses of #115*g# #AgNO_3(aq)# and #85*g# of #BaCl_2# where each of the components are dissolved in water?

1 Answer
May 30, 2017

Answer:

Approx. #100*g# #AgCl# precipitate should deposit........

Explanation:

First we need a stoichiometric equation........

#2AgNO_3(aq) + BaCl_2(aq) rarr 2AgCl(s)darr +Ba(NO_3)_2#

Now while nitrates, and most halides are soluble; silver chloride is as soluble as a brick, and precipitates from aqueous solution as a curdy white precipitate.

So we works out equivalent quantities of silver nitrate, and barium chloride.

#"Moles of silver nitrate"=(115*g)/(169.87*g*mol^-1)=0.677*mol.#

#"Moles of barium chloride"=(85*g)/(208.23*g*mol^-1)=0.408*mol# with respect to barium chloride; but #0.816*mol# with respect to chloride ion. Are you with me?

And thus, because chloride is in excess, we should achieve stoichiometric precipitation of silver chloride, i.e. #0.677*mol# #AgCl#.

And tis represents a mass of #0.677*molxx143.32*g*mol^-1#

#=??*g#