Final answer:
To calculate the mass of Ag2CrO4 precipitate formed, the number of moles of AgNO3 is determined and used to calculate moles of Ag2CrO4 using the stoichiometry of the reaction. The molar mass of Ag2CrO4 is then used to convert these moles to grams.
Step-by-step explanation:
The question involves using stoichiometry to calculate the mass of Ag2CrO4 that will precipitate from a silver nitrate solution when excess potassium chromate is added. The balanced chemical equation is:
2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq)
The volume and concentration of AgNO3 are given, so first, we calculate the moles of AgNO3:
moles of AgNO3 = 0.415 L * 0.186 mol/L = 0.07719 mol
According to the stoichiometry of the reaction, 2 moles of AgNO3 yield 1 mole of Ag2CrO4. Therefore, the moles of Ag2CrO4 that can be formed are:
moles of Ag2CrO4 = 0.07719 mol AgNO3 / 2 = 0.038595 mol
To find the mass of Ag2CrO4 formed, we need the molar mass of Ag2CrO4 (331.73 g/mol). The mass is calculated as follows:
mass of Ag2CrO4 = 0.038595 mol * 331.73 g/mol = 12.81 g
This mass does not match any of the options given (a. 46.1 g b. 92.2 g c. 184.4 g d. 368.8 g), likely indicating an error in the question or a misunderstanding of the values provided.