In a chemical analysis, a technician mixes silver nitrate solution with excess sodium chromate solution to produce 2.89 g of precipitate. What mass of silver nitrate was present in the original solution?

A chromium (III) chloride solution is analyzed by having a sample of the solution react with a 50.0 g piece of zinc metal. After the reaction, 38.5 g of zinc remained. Calculate the mass of chromium (III) chloride that was present in the sample tested.

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Originally Posted by Jessica106

In a chemical analysis, a technician mixes silver nitrate solution with excess sodium chromate solution to produce 2.89 g of precipitate. What mass of silver nitrate was present in the original solution?

A chromium (III) chloride solution is analyzed by having a sample of the solution react with a 50.0 g piece of zinc metal. After the reaction, 38.5 g of zinc remained. Calculate the mass of chromium (III) chloride that was present in the sample tested.

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Well first write out the chemical equation.
2AgNO3 + Na2CrO4 = Ag2CrO4 + 2NaNO3

The precipitate will be Silver Chromate as it is only "slightly soluble" while Sodium Nitrate is "soluble" (check a solubility table). Then you can calculate the mass of the Silver Nitrate from the mass of Silver Chromate (AgNO3 would be the limiting reagent - there is an excess of Sodium Chromate).

mass silver chromate * molar mass silver chromate = moles silver chromate * mole ratio = moles silver nitrate * molar mass silver nitrate = mass silver nitrate

2.89 * 1/331.73 = 0.00871190426 * 2/1 = 0.0174238085 * 169.8371/1 = 2.95920911

Therefore 2.96g of Silver Nitrate were present in the original solution. Hope this helps!

(I am a grade 11 chemistry student)
- Jason