In the reaction of iron(III) sulfate and barium hydroxide, Fe2(SO4)3 + 3 Ba(OH)2 3 BaSO­4 + 2 Fe(OH)3 If 20.0 g of Fe2(SO4)3 is mixed with 20.0 g of Ba(OH)2 , Molar Masses: Fe2(SO4)3 = 399.91 Ba(OH)2 = 171.35 Fe(OH)3 = 106.88
How many moles of Fe(OH)3 will be produced if all 20.0 g of Fe2(SO4)3 were consumed?
How many moles of Fe(OH)3 will be produced if all 20.0 g of Ba(OH)2were consumed?
How many grams of Fe(OH)3 will be produced if both iron(III) sulfate and barium hydroxide were used in the reaction?
Which reactant was the limiting reactant?
If one of my Chem 10 students ran the reaction as stated in problems 35 and got 7.052 g of Ba(OH)2, what is the percent yield for the reaction?