# Problem: Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 6H2O(l) Part A Which reagent is the limiting reactant when 0.450 mol Al(OH)3 and 0.450 mol H2SO4 are allowed to react? a) H2O(l) b) H2SO4(aq) c) Al(OH)3(s) d) Al2(SO4)3(aq)Part BHow many moles of Al2(SO4)3 can form under these conditions? Express the amount in moles to three significant digits. Part C How many moles of the excess reactant remain after the completion of the reaction? Express the amount in moles to three significant digits.

###### FREE Expert Solution

Reaction:

2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)

Part A

moles of Al2(SO4)3  from 0.450 mol Al(OH)3

0.225 mol Al2(SO4) ###### Problem Details

Aluminum hydroxide reacts with sulfuric acid as follows:

2Al(OH)3(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 6H2O(l)

Part A

Which reagent is the limiting reactant when 0.450 mol Al(OH)3 and 0.450 mol H2SO4 are allowed to react?

a) H2O(l)

b) H2SO4(aq)

c) Al(OH)3(s)

d) Al2(SO4)3(aq)

Part B

How many moles of Al2(SO4)3 can form under these conditions?

Express the amount in moles to three significant digits.

Part C

How many moles of the excess reactant remain after the completion of the reaction?

Express the amount in moles to three significant digits.