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....

$143.32*g·mol^-1xx0.129*mol-=??g$