What mass of Na2CrO4 is required to precipitate all the silver ions from 75.0 mL of .100M solution of AgNO3?

What mass of Na2CrO4 is required to precipitate all of the silver ions from 75.0 mL of a .100M solution of AgNO3?

Answer:
2AgNO3 + Na2CrO4 --> Ag2CrO4 + 2NaNO3

To determine how much Na2CrO4 is needed, you need to find the number of moles of AgNO3 that you have.

moles AgNO3 = molarity * volume
= 0.100 mol/L * 0.075 L
= 0.0075 mol AgNO3

looking at your balanced equation, you need 1 mol Na2CrO4 for every two moles of AgNO3.

So using a mole ratio of 1:2
0.0075 mol AgNO3 * 1 mol Na2CrO4 / 2 mol AgNO3
= 0.00375 mol Na2CrO4

Now to get the mass of Na2CrO4 needed,
mass = mol * molar mass
=0.00375mol * (2*23g/mol + 52g/mol + 4*16g/mol)
= 0.6075 g

Therefore to precipitate all the silver ions, you would need 0.608g of Na2CrO4.
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