What mass of Na2CrO4 is required to precipitate all the silver ions from 96.0 mL of a 0.650 M solution of AgNO3?
Here is the equation.
Na2CrO4 + 2Ag^+ --> Ag2CrO4 + 2Na^+
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
thank u dr bob
the computer is not accepting the answer, can u answer this and all the other posts i have
To solve this problem, we need to use stoichiometry to determine the amount of Na2CrO4 needed to precipitate all the silver ions.
First, let's write the balanced equation for the reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4):
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to form 1 mole of Ag2CrO4.
Step 1: Calculate the number of moles of AgNO3 in the solution.
To find the number of moles of AgNO3, we can use the formula:
moles = concentration x volume.
moles of AgNO3 = 0.650 M x (96.0 mL / 1000 mL/1 L)
moles of AgNO3 = 0.0624 moles
Step 2: Use the stoichiometry from the balanced equation to determine the number of moles of Na2CrO4 required.
From the balanced equation, we know that the molar ratio between AgNO3 and Na2CrO4 is 2:1.
moles of Na2CrO4 = 0.0624 moles AgNO3 x (1 mole Na2CrO4 / 2 moles AgNO3)
moles of Na2CrO4 = 0.0312 moles
Step 3: Calculate the mass of Na2CrO4 needed.
To find the mass of Na2CrO4, we can use the formula:
mass = moles x molar mass.
Molar mass of Na2CrO4 = (2 x atomic mass of Na) + atomic mass of Cr + (4 x atomic mass of O)
Molar mass of Na2CrO4 = (2 x 22.99 g/mol) + 52.00 g/mol + (4 x 16.00 g/mol)
Molar mass of Na2CrO4 = 161.97 g/mol
mass of Na2CrO4 = 0.0312 moles x 161.97 g/mol
mass of Na2CrO4 = 5.05 g
Therefore, approximately 5.05 grams of Na2CrO4 is required to precipitate all the silver ions from 96.0 mL of a 0.650 M solution of AgNO3.