Ch 202 Problem 3.72

Aluminum hydroxide reacts with sulfuric acid as follows:

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

Which reagent is the limiting reactant when 0.500 mol Al(OH)3 and 0.500 mol H2SO4 are allowed to react?

Limiting reactant problems are ALWAYS worked in terms of moles!!! If we're given gram amounts of reactants, we convert g -> mol and then proceed. Here, we're given reactant amounts in moles, so we just convert mol Al(OH)3 to mol H2SO4, and mol H2SO4 to mol Al(OH)3 and see which one we don't have enough of......

first, mol Al(OH)3 --> mol H2SO4:

we need 0.750 mol H2SO4 to consume 0.500 mol Al(OH)3. We only have 0.500 mol H2SO4, so H2SO4 is the limiting reactant!

verify by converting mol H2SO4 --> mol Al(OH)3:

we need 0.333 mol Al(OH)3 to consume 0.500 mol H2SO4. We have 0.500 mol Al(OH)3, so we're good!

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

we're asking for the theoretical yield of the aluminum sulfate here. To get the theoretical yield, we convert mol limiting reactant to mol product: here, this is mol H2SO4 --> mol Al2(SO4)3....

so we can form 0.167 mol aluminum sulfate.

how many mols of the excess reactant remain after completion of the reaction?

Al(OH)3 is the XS reactant, and we started with 0.500 mol. From above, we calculated how many mol Al(OH)3 were required to consume 0.500 mol H2SO4:

so, if we started with 0.500 mol Al(OH)3, and consumed 0.333 mol ,we must have

left.

DAC 7/3/08