An excess of aqueous AgNO3 reacts with
44.5 mL of 5 M K2CrO4(aq) to form a precip-
itate. What is the precipitate?
1. K2CrO4
2. KNO3
3. Ag2CrO4
4. AgNO3
What mass of precipitate is formed?
Answer in units of g
You need to know the solubility rules. Here is a simplified set.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
To determine the precipitate formed in the reaction between excess aqueous AgNO3 and 44.5 mL of 5 M K2CrO4(aq), we need to consider the reaction between AgNO3 and K2CrO4.
The balanced chemical equation for the reaction is:
2 AgNO3 + K2CrO4 → Ag2CrO4 + 2 KNO3
From the equation, we can see that Ag2CrO4 is the precipitate formed. So, the answer to the first part of the question is option 3. Ag2CrO4.
To find the mass of the precipitate formed, we need to use the given volume and concentration of K2CrO4(aq) to determine the number of moles of K2CrO4. Then, we can use the stoichiometry of the balanced equation to find the number of moles of Ag2CrO4 formed. Finally, we can convert the moles of Ag2CrO4 to grams using its molar mass.
Here's how to calculate the mass of precipitate formed:
1. Calculate the number of moles of K2CrO4:
Molarity (M) = moles/volume (L)
5 M = moles/0.0445 L (convert mL to L)
moles of K2CrO4 = 5 M × 0.0445 L
2. Use the stoichiometry of the balanced equation to determine the number of moles of Ag2CrO4 formed. From the equation, we can see that 2 moles of AgNO3 react with 1 mole of K2CrO4 to produce 1 mole of Ag2CrO4:
moles of Ag2CrO4 = (moles of K2CrO4) × (1 mole Ag2CrO4 / 1 mole K2CrO4)
3. Convert the moles of Ag2CrO4 to grams using its molar mass. The molar mass of Ag2CrO4 can be calculated as:
(2 × atomic mass of Ag) + atomic mass of Cr + (4 × atomic mass of O)
4. Calculate the mass of precipitate formed by multiplying the number of moles of Ag2CrO4 by its molar mass.
Once you have the calculated value, you can provide the answer to the second part of the question in units of grams.