How many milliliters of 0.645 M K2CrO4 are needed to precipitate all the silver in 405 mL of 0.190 M AgNO3 as Ag2CrO4?
2AgNO3 + K2Cr2O7 ==> 2KNO3 + Ag2CrO4
moles AgNO3 = M x L = ??
moles K2Cr2O7 = 1/2 x moles AgNO3
M K2Cr2O7 = moles/L
solve for L and convert to mL.
To determine the number of milliliters of 0.645 M K2CrO4 needed to precipitate all the silver in 405 mL of 0.190 M AgNO3 as Ag2CrO4, we can use a balanced chemical equation and stoichiometry.
The balanced chemical equation for the reaction between K2CrO4 and AgNO3 is:
2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of K2CrO4 to produce 1 mole of Ag2CrO4.
First, let's calculate the number of moles of silver nitrate (AgNO3) in 405 mL of 0.190 M AgNO3:
moles of AgNO3 = volume (in liters) × molarity
= 0.405 L × 0.190 mol/L
= 0.07695 moles
Since the stoichiometry of the reaction tells us that 2 moles of AgNO3 react with 1 mole of K2CrO4, we can determine the number of moles of K2CrO4 needed:
moles of K2CrO4 = (0.07695 moles AgNO3) / (2 moles AgNO3/1 mole K2CrO4)
= 0.038475 moles
Now, we need to determine the volume of 0.645 M K2CrO4 containing 0.038475 moles:
volume = moles / molarity
= 0.038475 moles / 0.645 mol/L
= 0.059695 L
Finally, we convert the volume in liters to milliliters:
volume = 0.059695 L × 1000 mL/L
= 59.695 mL
Therefore, approximately 59.695 milliliters of 0.645 M K2CrO4 are needed to precipitate all the silver in 405 mL of 0.190 M AgNO3 as Ag2CrO4.