Question

We wish to determine how many grams of solid silver chromate will precipitate when 150. mL of 0.500 M silver nitrate solution is added to excess potassium chromate.

2AgNO₃ (aq) + K2CrO₄ (aq) Ag₂CrO₄ (s) + AKNO₃ (aq).
In the previous step, you determined 0.075 AgNO₃ reaction. The molar mass of Ag₂CO₄ is 334.71 g/mol. How many grams of Ag₂CrO₄ form during the reaction?

Answer 1

To find the mass of silver chromate precipitated from the reaction of silver nitrate and potassium chromate, convert 0.075 moles of silver nitrate to moles of silver chromate using stoichiometry and then to grams using the molar mass, which results in 12.55 grams.

Step-by-step explanation:

To determine how many grams of solid silver chromate would precipitate when 150 mL of 0.500 M silver nitrate solution is mixed with excess potassium chromate, we can use stoichiometry.

The balanced chemical equation is 2AgNO₃ (aq) + K₂CrO₄ (aq) → Ag₂CrO₄ (s) + 2KNO₃ (aq). First, calculate the moles of AgNO₃ using its concentration and the volume of the solution:

• 0.500 moles/L * 0.150 L = 0.075 mol AgNO₃

According to the stoichiometry of the balanced chemical equation, 2 moles of AgNO₃ are needed to produce 1 mole of Ag₂CrO₄. Thus, 0.075 moles of AgNO₃ would produce 0.075/2 = 0.0375 moles of Ag₂CrO₄.

Now we convert the moles of Ag₂CrO₄ to grams using its molar mass:

• 0.0375 mol * 334.71 g/mol = 12.551625 grams of Ag₂CrO₄