Answered

Aqueous potassium chromate reacts with a solution of silver nitrate forming aqueous potassium nitrate and a solid silver chromate. If 0.0120 g of silver chromate is formed, what volume of a 0.50 M silver nitrate solution reacted?



Answer :

Answer:

0.111 mL

Explanation:

To solve this problem, you need to use stoichiometry and the concept of limiting reactant.

First, write the balanced equation for the reaction:

K2CrO4 (aq) + 2AgNO3 (aq) → 2KNO3 (aq) + Ag2CrO4 (s)

From the equation, you can see that the mole ratio of K2CrO4 to Ag2CrO4 is 1:1. This means that the amount of Ag2CrO4 formed is directly proportional to the amount of K2CrO4 reacted.

Next, calculate the number of moles of Ag2CrO4 formed:

moles of Ag2CrO4 = mass / molar mass

moles of Ag2CrO4 = 0.0120 g / (2 x 107.87 g/mol)

moles of Ag2CrO4 = 5.56 x 10^-5 mol

Since the mole ratio of K2CrO4 to Ag2CrO4 is 1:1, the number of moles of K2CrO4 reacted is also 5.56 x 10^-5 mol.

Now, you need to use the concentration of the silver nitrate solution to calculate the volume of solution used:

moles of AgNO3 = Molarity x Volume (in liters)

Volume (in liters) = moles of AgNO3 / Molarity

Volume (in liters) = 5.56 x 10^-5 mol / 0.50 mol/L

Volume (in liters) = 1.11 x 10^-4 L

Finally, convert the volume from liters to milliliters:

Volume (in mL) = 1.11 x 10^-4 L x 1000 mL/L

Volume (in mL) = 0.111 mL

Therefore, the volume of the 0.50 M silver nitrate solution that reacted is 0.111 mL.