You are correct that Al(OH)3 is the limiting reagent. I don't know how you calculated that but that is the right answer. After you know the limiting reagent (LR), the remainder of the problem starts there. You solve it as you would any stoichiometry problem.
2.4 moles Al(OH)3 x (3 moles H2SO4/2 moles Al(OH)3) = 2.4 x (3/2) = 3.6 moles H2SO4 used; therefore, 5.3-3.6 = moles remaining.
You could have used the Al2(SO4)3 to do it also.
1.2 moles Al2(SO4)3 x [3 moles H2SO4/1 mole Al2(SO4)3] = 1.2*3 = 3.6 moles H2SO4 used.
In a reaction vessel, 2.4 mol of Al(OH)3 and 5.3 mol of H2SO4 react.
Products: Al2(SO4)3, H2O
[I know...]
Moles of Al2(SO4)3 in container:1.2mol
Moles of H2O in container:7.2 mol
[But not...]
Moles of excess reactant in container:____mol
how do i find it?
1 answer
1 answer