# 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?

May 30, 2017

Approx. $100 \cdot g$ $A g C l$ precipitate should deposit........

#### Explanation:

First we need a stoichiometric equation........

$2 A g N {O}_{3} \left(a q\right) + B a C {l}_{2} \left(a q\right) \rightarrow 2 A g C l \left(s\right) \downarrow + B a {\left(N {O}_{3}\right)}_{2}$

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.

$\text{Moles of silver nitrate} = \frac{115 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1} = 0.677 \cdot m o l .$

$\text{Moles of barium chloride} = \frac{85 \cdot g}{208.23 \cdot g \cdot m o {l}^{-} 1} = 0.408 \cdot m o l$ with respect to barium chloride; but $0.816 \cdot m o l$ 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. $0.677 \cdot m o l$ $A g C l$.

And tis represents a mass of $0.677 \cdot m o l \times 143.32 \cdot g \cdot m o {l}^{-} 1$

=??*g