Question

In the reaction K₂CrO₄ (aq) + PbCl₂ (aq) -- 2KCl (aq) + PbCrO₄(s), how many grams of Pbcro, will precipitate out from the reaction between 25.0 milliers of 3.0 M K Cro, in an excess of PbCl?

A 75 grams
B. 0.024 grams
C. 24 grams
D. 075 grams
E. 3.0 grams

Answer 1

The mass of PbCrO4 precipitating from the reaction of 25.0 mL of 3.0 M K2CrO4 with an excess of PbCl2 is approximately 24 grams, corresponding to answer C.

Step-by-step explanation:

To calculate the mass of PbCrO4 that will precipitate out from the reaction given 25.0 milliliters of 3.0 M K2CrO4 with an excess of PbCl2, first we need to convert the volume of K2CrO4 to liters and then calculate the number of moles of K2CrO4. Using the stoichiometry of the reaction, we can find the moles of PbCrO4 precipitate. After that, we use the molar mass of PbCrO4 to convert the moles to grams. 25.0 mL K2CrO4 = 0.025 L K2CrO4
Moles of K2CrO4 = 0.025 L × 3.0 mol/L = 0.075 mol K2CrO4
Since the reaction is 1:1, moles of PbCrO4 will also be 0.075.
The molar mass of PbCrO4 is approximately 323.2 g/mol.

Mass of PbCrO4 = 0.075 mol × 323.2 g/mol = 24.24 grams

Therefore, the mass of PbCrO4 that will precipitate is approximately 24 grams when 25.0 mL of 3.0 M K2CrO4 reacts with an excess of PbCl2, which corresponds to answer C.