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 23, 2017

WE need (i) a stoichiometric equation........and should get over $12 \cdot g$ of a brick-red precipitate.

Explanation:

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

Silver chromate is EXCEPTIONALLY insoluble, and will deposit with alacrity as a brick-red precipitate.

And we need equivalent quantities of silver ion and chromate ion.

$\text{Moles of silver ion} = 150 \times {10}^{-} 3 \cdot L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.075 \cdot m o l$

$\text{Moles of chromate ion} = 100 \times {10}^{-} 3 \cdot L \times 0.400 \cdot m o l \cdot {L}^{-} 1 = 0.040 \cdot m o l$

Clearly, there is EXCESS chromate ion, MORE than 1/2 equiv. And thus all the silver ion will precipitate as silver chromate.

Given the stoichiometry, $\frac{0.075}{2} \cdot m o l$ $A {g}_{2} C r {O}_{4}$ (approx. 12*g) should precipitate......

i.e. 0.0375*molxx331.73*g*mol^-1=??*g