Question

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 0.300 mol Al(OH)3 and 0.300 mol H2SO4 are allowed to react? a) Al(OH)3(s) b) H2SO4(aq) c) Al2(SO4)3(aq) d) H2O(l)

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.

Answer 1

A)

Balanced chemical equation is:

2 Al(OH)3 + 3 H2SO4 ---> Al2(SO4)3 + 6 H2O

2 mol of Al(OH)3 reacts with 3 mol of H2SO4

for 0.3 mol of Al(OH)3, 0.45 mol of H2SO4 is required

But we have 0.3 mol of H2SO4

so, H2SO4 is limiting reagent

Answer: b

B)

we will use H2SO4 in further calculation

According to balanced equation

mol of Al2(SO4)3 formed = (1/3)* moles of H2SO4

= (1/3)*0.300

= 0.100 mol

Answer: 0.100 mol

C)

According to balanced equation

mol of Al(OH)3 reacted = (2/3)* moles of H2SO4

= (2/3)*0.300

= 0.200 mol

mol of Al(OH)3 remaining = mol initially present - mol reacted

mol of Al(OH)3 remaining = 0.300 - 0.200

mol of Al(OH)3 remaining = 0.100 mol

Answer: 0.100 mol