In: Chemistry
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 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