# What is the mass of precipitate formed when 50 ml of 16.9% w/v solution of AgNO3 is mixed with 50 ml of 5.8% w/v NaCl solution?

Aug 17, 2017

Well, $\text{w/v"-="Mass of solute"/"Volume of solution}$, we should get over $7 \cdot g$ of silver chloride.

#### Explanation:

And thus "mass of silver nitrate"=16.9%xx50*mL=8.45*g.

And this represents a molar quantity of $\frac{8.45 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$

$= 0.0497 \cdot m o l$, with respect to $A g N {O}_{3}$.

Likewise "mass of sodium chloride"=5.8%xx50*mL=2.90*g.

And this represents a molar quantity of $\frac{2.90 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1}$

$= 0.0497 \cdot m o l$ with respect to $N a C l$.

Clearly the reagents are present in 1:1 molar ratio. The reaction that occurs in solution is the precipitation of a curdy white mass of $A g C l \left(s\right)$, i.e. and we represent this reaction by the net ionic equation.....

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

Of course the complete reaction is....

$A g N {O}_{3} \left(a q\right) + N a C l \left(a q\right) \rightarrow A g C l \left(s\right) \downarrow + N a N {O}_{3} \left(a q\right)$, i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.

And given the stoichiometry, we gets $0.04974 \cdot m o l \times 143.32 \cdot g \cdot m o {l}^{-} 1 = 7.11 \cdot g$.

Of course a material such as silver halide would be very hard to isolate. Particle size is very small; it is likely to clog the filter, and filter very slowly; and moreover $A g C l$ is photoactive, and would decompose under light.