CHEM Assignment 24:  Stoichiometry #4 - Limiting Reactants 

Questions 1-3 are all based on this balanced equation. Molecular weights are given here! 

   2Al(OH)3     +       3H2SO4        Al2(SO4)3     +      6H2O

   MW 78.00 MW = 98.08       MW =342.15 MW =18.02

1.   In the 1st experiment, Precisely 10.00 g of aluminum hydroxide is stirred into 25.0 mL of 6.00 M sulfuric acid.  What mass of aluminum sulfate can be formed?

                    x                              x                              x                              =

                    x                              x                              x                              =

   2Al(OH)3     +       3H2SO4        Al2(SO4)3     +      6H2O

2.   In a 2nd experiment, 5.00 g of Al(OH)3 are reacted with 9.00 g of H2SO4. Show by calculations which is the limiting reactant and determine how many grams of the excess reactant are left over?

                    x                              x                              x                              =

                    x                              x                              x                              =

2Al(OH)3 +  3H2SO4   Al2(SO4)3 + 6H2O

3.   In a 3rd experiment, 75.0 mL of 2.50 M sulfuric acid are reacted with 1.95 g of aluminum hydroxide.  Assume the solution volume remains at 75.0 mL. What is the molarity of the sulfuric acid after the reaction? (The nature of this question actually tells you which chemical is the limiting reactant, right? - Solve for mol/L (M) sulfuric acid needed! If you have the molarity given and you know the amount you need out of that....how would you determine the amount left?)

                    x                              x                              x                              =

2C8H18   +     25O2         16CO2   +    18H2O

                   MW 114.2         MW = 32.00

4.   If 2.58 x 104 g of octane are reacted with 8.75 x104 g of oxygen, which reactant is in excess and how many grams of it are left over?

                    x                              x                              x                              =

                    x                              x                              x                              =

2Bi    +     3H2       2BiH3

5.  If 10.00 grams of pure hydrogen is reacted with 650.0 g of bismuth, bismuth(III) hydride is formed. Determine which reactant is in excess and calculate the mass of the excess reactant.

                    x                              x                              x                              =

                    x                              x                              x                              =

2KOH     +    H2SO4         K2SO4    +    2H2O

6.   In a 3rd experiment, precisely 50.0 mL of 3.00 M sulfuric acid is mixed with 400.0 mL of 1.00 M KOH. What chemical is left over and what is its final molarity? It may be useful to solve just for MOLES needed - so only two step calcs to start. Think about moles of each that are given (V x M) and then how many moles are needed. (And this added hint: you're pouring one solution into the other....what is happening to the volume?)

              x                              x                              =

                    x                              x                              =