# How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate?

Apr 27, 2017

We interrogate the stoichiometric equation:

$2 A {g}^{+} + C r {O}_{4}^{2 -} \rightarrow A {g}_{2} C r {O}_{4} \left(s\right) \downarrow$

#### Explanation:

$\text{Silver chromate}$ will deposit from solution as a fine brick-red precipitate with alacrity; the material is exceptionally insoluble.

And thus we calculate the equivalent quantities of $\text{silver ion}$ and $\text{chromate ion}$.

$\text{Moles of silver nitrate} = 0.150 \cdot L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.0750 \cdot m o l$.

$\text{Moles of potassium chromate} = 0.100 \cdot L \times 0.400 \cdot m o l \cdot {L}^{-} 1 = 0.0400 \cdot m o l$.

Clearly, silver nitrate is the limiting reagent, as there is a stoichiometric excess of chromate ion.

And thus we should get a mass of 1/2xx0.0750*molxx 331.73*g*mol^-1=12.4*g with respect to $A {g}_{2} C r {O}_{4}$.