Question

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?

Answer 1

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`