Question

Mixing 39.0g of Al(OH)3 with an excess of H2SO4 we produce Al2(SO4)3. Using the following equation answer the following:

I have already balanced the equation from Al(OH)3+H2SO4---->Al2(SO4)3+H2O

2Al(OH)3 + 3H2504 → Al2(SO4)3 + 6H2O a. Determine and circle the limiting reagent for A12(SO4)3 (show all its calculations). b. Determine and circle the yielding theory for Al2(S04)3 (show all its calculations). of Al2(SO4)3 (show all its Determine the % of yield, if the experimental yield is 73.25 calculations). c.

Answer 1

2 Al (OH)_3 + 3H_2 SO_4 rightarrow Al_2 (So_4)_3 + 6H_2O it is already mentioned in the question that H_2SO_4 is in excess amount this Al (OH)_3 is the limiting reagent to determine yield of Al_2 (SO_4)_3: grams product = grams reactant times (1 mol reactant/molar mass of reactant) times (mole ratio product/reactant) times (molar mass of product/1 mol product) mass of Al_2 (so_4)_3 = 39.0 g times (1 mol Al (OH_3) 78 g Al (OH)_3) times (1 mol Al_2 (SO_4)_3/2 mole Al (oH)_3) times (342.15 g Al_2 (SO_4)_3/1 mol Al_2 (SO_4)_3) All the units except, grams of Al_2 (SO_4)_3 cancel out mass of Al_2 (SO_4)_3 (g) = 39 times 342.15/78 times 2 = 85.53 g % yield = Actual yield/theoretical yield times 100 % = 73.25/85.52 (given) times 100 % = 85.64 %