Question

Answer the following questions based on the reaction: K2CrO4 + AgNO3 ---> red solid. A: What is the formula for the solid produced? B: Using the reaction above, a chemestry student has 35 mL of a .250 M solution of silver nitrate. How many milliliters of .105 M K2CrO4 will be required to react completely with the silver ion present. C: Considering the reaction in part b, how many grams of product will be formed?

Answer 1

Answer the following questions based on the reaction: K2CrO4 + AgNO3 ---> red solid.

A: What is the formula for the solid produced?

2 AgNO3 + K2CrO4 ---> Ag2CrO4 + 2 KNO3

the formula for the solid produced in this reaction is Ag2CrO4; silver chromate

B: Using the reaction above, a chemestry student has 35 mL of a .250 M solution of silver nitrate. How many milliliters of .105 M K2CrO4 will be required to react completely with the silver ion present.

Moles of AgNO3 = molarity * volume in L

= 0.250 M * 0.035 L

= 8.75*10^-3 moles AgNO3

Now calculate the moles of K2CrO4:

8.75*10^-3 moles AgNO3 * 1 mole K2CrO4/ 2 moles AgNO3

= 4.375*10^-3 moles K2CrO4

Molarity = number of moles / volume in L

Volume in L = number of moles / molarity

= 4.375*10^-3 moles K2CrO4/0.105 M K2CrO4

= 0.0416 L

= 41.6 mL

C: Considering the reaction in part b, how many grams of product will be formed?

2 AgNO3 + K2CrO4 ---> Ag2CrO4 + 2 KNO3

Moles of AgNO3 = molarity * volume in L

= 0.250 M * 0.035 L

= 8.75*10^-3 moles AgNO3

Now calculate the moles of Ag2CrO4

8.75*10^-3 moles AgNO3 * 1 Moles of Ag2CrO4/ 2 moles AgNO3

= 4.375*10^-3 moles Ag2CrO4

Amount of Ag2CrO4

= number of moles * molar mass

= 4.375*10^-3 moles Ag2CrO4* 331.73 g/mol

= 1.45 g