How many grams of silver chromate will precipitate when 150. mL of .500 M silver nitrate are added to 100. mL of .400 M of potassium chromate

2Ag^+ +CrO2/4→Ag2CrO4(s)↓

Explanation:

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

silver ion

and

chromate ion.

Moles of silver nitrate =0.150 ⋅ L × 0.500 ⋅ mol ⋅ L^−1 = 0.0750 ⋅ mol . Moles of potassium chromate = 0.100 ⋅ L × 0.400 ⋅ mol ⋅ L^−1 = 0.0400 ⋅ mol

.

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

And thus we should get a mass of

1/2 × 0.0750 ⋅ mol × 331.73 ⋅ g ⋅ mol^−1 = 12.4 ⋅ g

with respect to Ag2CrO4.

https://api-project-1022638073839.appspot.com/questions/how-many-grams-of-silver-chromate-will-precipitate-when-150-ml-of-0-500-m-silver#413727

the right answer is c i just did the question.