Find the mass of silver nitrate 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.
In the reaction 2AgNO3 (aq) + K2CrO4 (aq) --> Ag2CrO4 (s) + 2KNO3 (aq), one mole of silver chromate is formed as a product when 2 moles of silver nitrate react with potassium chromate. 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 silver nitrate is 169.87 g/mole and the molar mass of silver chromate is 331.73 g/mole.
0.778g of silver chromate is equivalent to 0.778/331.73 moles of silver chromate. To form this twice the number of moles of silver nitrate is required. This gives the number of moles of silver nitrate that reacts as 2*(0.778/331.73). Using the molar mass of silver nitrate 2*(0.778/331.73) moles have a mass of 2*(0.778/331.73)*169.87 = 0.7976 g
The mass of silver nitrate that reacted is 0.7976 g.