Mass of aluminum sulfate hydrate+ cup b Mass of empty cup 2 Moles f aluminum sul

Question

Mass of aluminum sulfate hydrate+ cup b Mass of empty cup 2 Moles f aluminum sulfate hydrate of aluminum sulfate hydrate (Note: the mass of 18H20 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 6 LIMITING REACTANT 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 Calculation

Explanation / Answer

3BaCl2(aq) + Al2(SO4)3(aq) 3BaSO4(s) + 2AlCl3(aq)

Moles of Al2(SO4)3 = Mass /Molar mass =0.8 / 666.425 = 1.2 milli moles

Moles of BaCl2 = Molarity*V in Litres = 10*0.5 /1000 = 5 milli moles

3 moles of BaCl2 reacts with 1 moles of Al2(SO4)3 and forms 3 moles of BaSO4

1.2 millimoles of Al2(SO4)3 .reacts with 1.2*3 = 3.6 millimoles of BaCl2 and forms 3.6 millimoles of BaSO4

Limiting reagent :

Here Al2(SO4)3.18H2O completely consumed , so the limiting reagent is Al2(SO4)3.18H2O

Mass of BaSO4 formed = 3.6*10^-3 *Molar mass = 233.38 = 0.84 gms

% Percent yield = Actual yield*100 / Theoritical yield

We have not percent yield ...this is the procedure for calculating the above given things

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