Question

We wish to determine how many grams of silver

chromate precipitate when 150. mL of 0.500 M
silver nitrate solution is added to excess
potassium chromate.
+ K₂ CrO4 (aq) → Ag2 CrO4(s) + 2KNO3(aq)
2AgNO3(aq)
In the previous step, you determined
0.075 mol AgNO3 react. The molar mass of
Ag₂ CrO4 is 334.74 g/mol.
How many grams of Ag₂CrO4 form during the
reaction?
Mass (g) Ag, CrO
Enter

Answer 1

To determine the mass of Ag₂CrO4 formed during the reaction, we can use the volume and concentration of AgNO3 solution to calculate the number of moles of AgNO3. From the balanced chemical equation, we know that the stoichiometric ratio between AgNO3 and Ag₂CrO4 is 2:1. By converting the moles of Ag₂CrO4 to grams using its molar mass, we find that the mass of Ag₂CrO4 formed is approximately 12.53 grams.

Step-by-step explanation:

To determine the mass of Ag₂CrO4 formed during the reaction, we need to first calculate the number of moles of AgNO3 using its volume and concentration. Given that 150 mL of 0.500 M AgNO3 solution is added, we can calculate the moles of AgNO3 as follows:

Moles of AgNO3 = Volume (L) x Concentration (M) = 0.150 L x 0.500 M = 0.075 mol AgNO3

Since the balanced chemical equation tells us that the stoichiometric ratio between AgNO3 and Ag₂CrO4 is 2:1, we know that 0.075 mol of AgNO3 will result in the formation of half the molar amount of Ag₂CrO4:

Moles of Ag₂CrO4 = 0.075 mol AgNO3 / 2 = 0.0375 mol Ag₂CrO4

Finally, we can convert the moles of Ag₂CrO4 to grams using its molar mass:

Mass of Ag₂CrO4 = Moles of Ag₂CrO4 x Molar mass = 0.0375 mol x 334.74 g/mol = 12.53 g