STOICHIOMETRY/LIMITING REAGENTS CHM 1045UCHM 1046L DATA SHEET NAME DATE STD.# SE

Question

STOICHIOMETRY/LIMITING REAGENTS CHM 1045UCHM 1046L DATA SHEET NAME DATE STD.# SEQUENCE # 1a Mass of aluminum sulfate hydrate + cup 1b Mass of empty cup 1c Mass of aluminum sulfate hydrate 2 Moles of aluminum sulfate hydrate (Note: the mass of the 18H2Os must be included in the molar mass) Calculation 3 Volume of barium chloride solution 4 Concentration of barium chloride solution (label) 5 Moles of barium chloride used Calculation: 6 LIMITING REAGENT Balanced equation Clearly demonstrate which is the limiting reagent Calculation for theoretical yield 7 Theoretical yield of barium sulfate 8 Mass of empty filter paper 9 Mass of filter paper+ dry precipitate 10 Mass of barium sulfate 11 Percent yield 0.9 Calculation

Explanation / Answer

Ans. #2. Moles of Al2(SO4)3.18H2O = Mass / Molar mass

                                                = 0.8 g / (666.428918 g/mol)

                                                = 0.00120043 mol

#5. Moles of BaCl2 = Molarity x Volume of solution in liters

                                                = 0.50 M x 0.010 L

                                                = 0.005 mol

#6. Balanced reaction:          3 BaCl2 + Al2(SO4)3 = 3 BaSO4 + 2 AlCl3

Theoretical molar ratio of reactants = 3 BaCl2 + Al2(SO4)3 = 1 : 3

Experimental molar ratio of reactants = 3 BaCl2 + Al2(SO4)3

                                                = 0.005 mol : 0.00120043 mol

                                                = 4.17 : 1

Comparing the theoretical and experimental molar ratio of reactants, the moles of BaCl2 is greater than 3.0 while that of Al2(SO4)3 is kept constant at 1.0 mol

So-

BaCl2 is the reagent in excess

Al2(SO4)3 is the limiting reactant.

# The formation of product follows the stoichiometry of limiting reactants.

According to the stoichiometry of balanced reaction, 1 mol Al2(SO4)3 produces 3 mol BaSO4.

So,

Theoretical moles of BaSO4 produced = 3 x Moles of Al2(SO4)3 consumed

                                                = 3 x 0.00120043 mol

                                                = 0.00360129 mol

Theoretical mass of BaSO4 produced = Theoretical moles x Molar mass

                                                = 0.00360129 mol x (233.3906 g/ mol)

                                                = 0.8405 g

Hence, theoretical yield of BaSO4 = 0.8405 g

#11. % yield = (Mass of BaSO4 / Theoretical yield) x 100

                        = (1.08 g / 0.8405 g) x 100

                        = 128.49 %

Note: The theoretical yield of reaction must be equal to or less than 100.0%.

A % yield greater than 100.0 % means experimental errors like partial drying of precipitate, etc.

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