Find the mass of potassium chromate that reacted.
Solutions of silver nitrate and potassium chromate react to produce a red precipitate of silver chromate:
2AgNO3 (aq) + K2CrO4 (aq) à Ag2CrO4 (s) + 2KNO3 (aq)
0.778 g of precipitate is formed in the reaction.
1 Answer | Add Yours
In the reaction 2AgNO3 (aq) + K2CrO4 (aq) --> Ag2CrO4 (s) + 2KNO3 (aq), one mole of silver chromate is formed as a product when 1 mole of potassium chromate reacts with two moles of silver nitrate. The mass of the precipitate formed when a certain quantity of silver nitrate reacts with potassium chromate is equal to 0.778 g.
The molar mass of potassium chromate is 194.19 g/mole and the molar mass of silver chromate is 331.73 g/mole.
0.778 g of silver chromate is equivalent to 0.778/331.73 moles of silver chromate. To form this, an equal number of moles of potassium chromate are required. This gives the number of moles of potassium chromate that reacts as (0.778/331.73). Using the molar mass of potassium chromate that is 194.19 g/mole, (0.778/331.73) moles have a mass of (0.778/331.73)*194.19 g = 0.4558 g
The mass of potassium chromate that reacted is 0.4558 g.
We’ve answered 317,596 questions. We can answer yours, too.Ask a question