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

Jul 19, 2017

Under $20 \cdot g$ of the halide may be precipitated.....

#### Explanation:

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

$A {g}^{+} + C {l}^{-} \rightarrow A g C l \left(s\right) \downarrow$

And thus moles of $A {g}^{+} \equiv A g$

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

$1.09 \cdot L \times 0.118 \cdot m o l \cdot {L}^{-} 1 = 0.129 \cdot m o l$ with respect to $A {g}^{+}$.

And thus an equimolar quantity of $A g C l$ may be precipitated, which is as soluble as a brick (but much more difficult to handle!).

And thus mass of $\text{silver chloride}$ possible is....

143.32*g·mol^-1xx0.129*mol-=??g