An excess of aqueous AgNO3 reacts with
24.5 mL of 5 M K2CrO4(aq) to form a precipitate. What is the precipitate?
Answer: Ag2CrO4
What mass of precipitate is formed?
Answer in units of g.
Yes, the ppt is Ag2CrO4.
Here is a link for a step by step procedure for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of precipitate that is formed, we need to determine the limiting reagent in the reaction and then use stoichiometry to find the corresponding mass of the precipitate.
First, let's determine the limiting reagent.
We have the following reaction:
2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq)
From the given information, we have 24.5 mL of 5 M K2CrO4(aq). To determine the limiting reagent, we need to convert the volume (mL) of K2CrO4(aq) to moles.
Firstly, let's convert the volume (mL) to liters (L):
24.5 mL * (1 L / 1000 mL) = 0.0245 L
Next, we can calculate the number of moles of K2CrO4 using the given molarity:
moles of K2CrO4 = volume (L) * molarity
moles of K2CrO4 = 0.0245 L * 5 M = 0.1225 moles
Now, let's use the stoichiometry of the balanced chemical equation to determine the moles of Ag2CrO4 that can be formed.
From the balanced equation, we see that the ratio between K2CrO4 and Ag2CrO4 is 1:1. Therefore, the number of moles of Ag2CrO4 is also 0.1225 moles.
We can then convert the moles of Ag2CrO4 to grams using the molar mass of Ag2CrO4.
The molar mass of Ag2CrO4 can be calculated as follows:
2(Ag) + (Cr) + 4(O) = 2(107.87 g/mol) + 52 g/mol + 4(16.00 g/mol) = 331.87 g/mol
Finally, we can calculate the mass of precipitate formed using the moles and molar mass:
mass of Ag2CrO4 = moles of Ag2CrO4 * molar mass of Ag2CrO4
mass of Ag2CrO4 = 0.1225 moles * 331.87 g/mol
mass of Ag2CrO4 ≈ 40.6 g
Therefore, the mass of the precipitate formed, Ag2CrO4, is approximately 40.6 grams.