Question

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

Answer 1

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

Explanation:

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`