Chemistry, 24.04.2020 21:33 ruby562

Joe Student mixes 20.0 ml of a 3.00 M solution of silver nitrate with 15.0 ml of a 2.00 M solution of sodium phosphate. He collects and dries the precipitate formed, then weighs the precipitate and finds that it has a mass of 8.00 grams. The precipitate is Silver phosphate. Answer the following questions based on this information.

A. Write the balanced equation for this reaction.
3 AgNO3(aq)+ Na3PO4(aq)Ag3PO4(ppt.)+ 3 NaNO3(aq)

B. How many moles of each reactant were there?
(0.0200 L)(3.00 M) = 0.0600 mol AgNO3
(0.0150 L)(2.00 M) = 0.0300 mol Na3PO4

C. What is the limiting reagent?
0.0300 mol Na3PO4(3 mol AgNO3/ 1 mol Na3PO4) = 0.0900 mol AgNO3neededonly have 0.0600 mol AgNO3so it is the limiting reagent

D. How many moles of excess reagent were left unreacted?
0.0600 mol L. R. (1 mol E. R. / 3 mol L. R.) = 0.0200 mol
E. R. needed0.0300 mol - 0.0200 mol = 0.0100 mol xcs

E. What is the precipitate?
Ag3PO4(ppt.)

F. How many moles of precipitate should have been formed theoretically?

Answers: 1

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Joe Student mixes 20.0 ml of a 3.00 M solution of silver nitrate with 15.0 ml of a 2.00 M solution o...