3.1 mol Fe(OH)3 and 7.5 mol H2SO4 react according to the equation 2 Fe(OH)3 + 3H2SO4 = Fe2(SO4)3 + 6H2O. If the limiting reactant is Fe(OH)3, calculate the amount of Fe2(SO4)3 formed.
I don't understand this question. If someone could show me how to get the question, that would be great.
I don't understand this question. If someone could show me how to get the question, that would be great.
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Actually the question is a bit messy. As the equation tells you that each mol of Fe(OH)3 needs 1.5 mol H2SO4 any you have got more than twice as much the sulphuric acid is clearly in excess so you don't need to be told that the Fe(OH)3 is the limiting reagent. (someone just trying to be helpful).
The limiting reagent is the one NOT in excess - it will be fully used up whereas other things will have some left over. (to be pedantic only 4.65 mol H2SO4 can actually react even if 7.5 mol was added)
So 2 moles Fe(OH)3 make 1 mole Fe2(SO4)3
so 3.1 mol makes 3.1/2= 1.55mol iron(III) sulphate.
The limiting reagent is the one NOT in excess - it will be fully used up whereas other things will have some left over. (to be pedantic only 4.65 mol H2SO4 can actually react even if 7.5 mol was added)
So 2 moles Fe(OH)3 make 1 mole Fe2(SO4)3
so 3.1 mol makes 3.1/2= 1.55mol iron(III) sulphate.