An excess of aqueous AgNo3 reacts with 41 mL of 5 M K2CrO4(aq) to form a precipitate. What is the precipitate?
Silver chromate--
Ag2CrO4
To determine the precipitate formed when an excess aqueous AgNO3 reacts with 41 mL of 5 M K2CrO4(aq), we need to consider the following reaction:
2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq)
In this reaction, silver nitrate (AgNO3) reacts with potassium chromate (K2CrO4) to form silver chromate (Ag2CrO4) precipitate and potassium nitrate (KNO3) as the soluble product.
Since we have an excess of AgNO3, it means that all the K2CrO4 will react, forming the maximum amount of Ag2CrO4 precipitate.
Now, let's calculate the amount of Ag2CrO4 precipitate formed:
First, let's convert the volume of K2CrO4 solution to moles using the concentration:
Volume = 41 mL = 41/1000 L = 0.041 L
Moles of K2CrO4 = Concentration × Volume = 5 M × 0.041 L = 0.205 mol
According to the balanced equation, 2 moles of AgNO3 react with 1 mole of K2CrO4 to form 1 mole of Ag2CrO4. Therefore, the amount of Ag2CrO4 precipitate formed will be half the moles of K2CrO4 used:
Moles of Ag2CrO4 = 0.205 mol / 2 = 0.1025 mol
Now, we can determine the molar mass of Ag2CrO4:
Ag2CrO4 = (2 × atomic mass of Ag) + atomic mass of Cr + (4 × atomic mass of O)
= (2 × 107.87 g/mol) + 52.00 g/mol + (4 × 16.00 g/mol)
= 331.74 g/mol
Finally, we can calculate the mass of Ag2CrO4 formed:
Mass = Moles × Molar mass = 0.1025 mol × 331.74 g/mol ≈ 34.01 g
Therefore, the precipitate formed is approximately 34.01 grams of silver chromate (Ag2CrO4).