What mass of silver chloride can be produced from 1.09 L of a 0.118 M solution of silver nitrate?

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Answer 1 · 2017-07-19T17:33:32

Under #20*g# of the halide may be precipitated.....

We want to perform a precipitation reaction........

#Ag^(+) + Cl^(-)rarrAgCl(s)darr#

And thus moles of #Ag^(+)-=Ag#

We gots a molar quantity of .................

#1.09*Lxx0.118*mol*L^-1=0.129*mol# with respect to #Ag^+#.

And thus an equimolar quantity of #AgCl# may be precipitated, which is as soluble as a brick (but much more difficult to handle!).

And thus mass of #"silver chloride"# possible is....

#143.32*g·mol^-1xx0.129*mol-=??g#