Aluminum hydroxide reacts with sulfuric acid as follows:Al(OH)3 + 3H2SO4--> ...?

Question

Aluminum hydroxide reacts with sulfuric acid as follows:
2Al(OH)3 + 3H2SO4--> Al2(SO4)3+6H2O
Which is the limiting reactant when .500 mol Al(OH)3 and .500 mol H2SO4 are allowed to react?
How many moles of Al2(SO4)3 can form under these conditions?
How many moles of the excess reactant remain after the reaction?

Answer 1

the ratio between Al(OH)3 and H2SO4 is 2 : 3 so H2SO4 is the limiting reactant

2 : 3 = x : 0.500

x = 0.333 moles Al2(SO4)3 needed for the reaction

0.500 - 0.333 = 0.167 moles Al2(SO4)3 in excess

the ratio between H2SO4 (limiting reactant) and Al2(SO4)3 is 3 . 1

so we get 0.500 / 30 = 0.167 moles Al2(SO4)3

Answer 2

2Al(OH)3+3H2SO4---> Al2SO4+6H20
You see that 2 mole of Al(OH)3 react with 3 moles H2SO4
So H2SO4 is the limitant reactant
IWith 0.5 mole of H2SO4 only 0.5*2/3=1/3=0.33mole of Al(OH)3 reacts and 1/2*1/3 = 1/6 mole of Al2(SO4)3 are formed
0.1167moles are formed
and in the solution 1/6 of Al(OH)3 remains

Answer 3

A, H2SO4 Since you have the same amount of both reactant, the one that is consumed more rapidly is the limiting reagent. If 0.5mol of AL(OH)3 were to react, you would need 0.75mol of H2SO4 (0.5*3/2). Hope that helps

Answer 4

FROM EQN, 2 mol Al2(oh)3 EQUIVALENT TO 3 MOL H2SO4
THUS, 0.5 mol Al2(oh)3 EQUIVALENT TO 0.75 MOL H2SO4
OR 0.33 mol Al2(oh)3 EQUIVALENT TO 0.5 MOL H2SO4
HENCE, Al2(oh)3 IS IN EXCESS AND H2SO4 IS THE LIMITING REAGENT
ALSO, FROM ABOVE EQN, 2 mol Al2(oh)3 EQUIVALENT TO 1 MOL AL2(SO4)3
THUS, 0.15 MOL AL2(SO4)3 IS FORMED.
ALSO, EXCESS REACTANT = 0.2 MOL Al2(oh)3