Stoichiometry practice...

Friday in class we looked at a problem set to practice stoichiometry. It's copied below. There is a new MasteringChemistry assignment posted, due Friday. Next week we will look at gas laws and continue with stoichiometry. Your next exam is October 5th.

Chem 150 – Fall 2009 – Problem Set #2

You are studying the reaction of 1.132M potassium phosphate solution with 1.275M barium nitrate solution.

1. Write a balanced chemical equation for this reaction.

a. How many grams of precipitate could you make if you completely react 125.0mL of the potassium phosphate solution?

b. How many grams of precipitate could you make if you completely react 175.0mL of the barium nitrate solution?

c. How many mL of the barium nitrate solution is required to react completely with 125.0mL of the potassium phosphate solution?

c. How many mL of the potassium phosphate solution is required to react completely with 175.0mL of the barium nitrate solution?

d. What is the theoretical yield of precipitate (in grams) if you react 125.0mL of the potassium phosphate solution with 175.0mL of the barium nitrate solution?

e. What is the limiting reagent in part d? How many moles of the excess reagent remain after the reaction is complete?

f. Write the balanced net ionic equation for this chemical process.

2. These questions deal with concentrations of the solutions used above:

a. What is the concentration of potassium ions in the 1.132M stock solution? Phosphate ions?

b. What is the concentration of barium ions in the 1.434M stock solution? Nitrate ions?

c. How many grams of potassium phosphate are present in 125.0mL of the 1.132M stock solution? How many grams of barium nitrate are present in 175.0mL of the 1.275M stock solution?

d. If you dissolved the mass of potassium phosphate in part c in enough water to make 300.0mL of solution, what would the concentration be? If you dissolved the mass of barium nitrate in part c in enough water to make 300.0mL of solution, what would the concentration be?

e. If you dilute 125.0mL of the potassium phosphate stock solution to a total volume of 300.0mL, what will be the “new” concentration of potassium phosphate? If you dilute 175.0mL of the barium nitrate stock solution to a total volume of 300.0mL, what will be the “new” concentration of barium nitrate?