Aluminum hydroxide reacts with sulfuric acid as follows: 2 al(oh)3(s) + 3 h2so4(aq) → al2(so4)3(aq) + 6 h2o(l). which reagent is the limiting reactant when 0.410 mol al(oh)3 and 0.410 mol h2so4 are allowed to react? h2so4 al2(so4)3 h2o al(oh)3 correct: your answer is correct. how many moles of al2(so4)3 can form under these conditions? 0.137 correct: your answer is correct. mol how many moles of the excess reactant remain after the completion of the reaction?

Balanced chemical equation for reaction of aluminium hydroxide and sulfuric acid is as follows:

(a) From the balanced chemical reaction, 2 mole of gives 1 mole of thus, 1 mole of gives 0.5 moles of and 0.410 moles of gives 0.205 moles of .

Molar mass of is 342.1509 g/mol thus, mass of will be:

m=n×M=0.205 mol×342.1509 g/mol=70.14 g

Similarly, 3 moles of gives 1 mole of thus, 1 mole of gives 0.33 moles of and 0.410 moles of  gives 0.136 moles of .

Thus, mass of will be:

m=n×M=0.136 mol×342.1509 g/mol=46.76g

Since, mass of    obtained from  is less than that from , is the limiting reactant.

(b) Number of moles of is 0.410 mol, from the balanced chemical reaction 3 moles of gives 1 mole of thus, 1 mole of gives 0.33 moles of and 0.410 moles of  gives 0.136 moles of .

Therefore, number of moles of formed is 0.136 mol.

(c) After the completion of reaction, all of the limiting reactant gets converted into product thus, 0.410 mole of reacts completely to give 0.136 mol of .

From the balanced chemical reaction, 1 mole of formed from 2 moles of , thus, 0.136 mol of formed from 0.136×2=0.272 mol.

Thus, 0.272 moles of is used and remaining moles will be (0.410-0.272) mol=0.138 mol.

Therefore, 0.138 mol of excess reactant remains after the completion of reaction.