Asked by Jill
A technician mixes a solution containing silver nitrate with a solution containing sodium chromate. 2.89 g of precipitate is produced (assume silver chromate). Calculate the mass of silver nitrate present in the first solution. Make sure you have the correct number of significant digits and units.
Answers
Answered by
DrBob222
You really don't know from the way the problem is stated; however, I assume the problem means "if the exact amount of reagents are mixed to produce the 2.89 g Ag2C4O4 and neither reagent is left".
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3.
mols Ag2CrO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Ag2CrO4 to mols AgNO3.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3.
mols Ag2CrO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Ag2CrO4 to mols AgNO3.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
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