Question

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?

Answer 1

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

$2Al(OH)_{3}(s)+3H_{2}SO_{4}(aq)\rightarrow Al_{2}(SO_{4})_{3}(aq)+6H_{2}O(l)$

(a) From the balanced chemical reaction, 2 mole of $Al(OH)_{3}$ gives 1 mole of $Al_{2}(SO_{4})_{3}$ thus, 1 mole of $Al(OH)_{3}$gives 0.5 moles of $Al_{2}(SO_{4})_{3}$ and 0.410 moles of $Al(OH)_{3}$gives 0.205 moles of $Al_{2}(SO_{4})_{3}$ .

Molar mass of $Al_{2}(SO_{4})_{3}$ is 342.1509 g/mol thus, mass of $Al_{2}(SO_{4})_{3}$ will be:

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

Similarly, 3 moles of $H_{2}SO_{4}$ gives 1 mole of $Al_{2}(SO_{4})_{3}$ thus, 1 mole of $H_{2}SO_{4}$ gives 0.33 moles of $Al_{2}(SO_{4})_{3}$ and 0.410 moles of $H_{2}SO_{4}$  gives 0.136 moles of $Al_{2}(SO_{4})_{3}$.

Thus, mass of $Al_{2}(SO_{4})_{3}$ will be:

m=n×M=0.136 mol×342.1509 g/mol=46.76g

Since, mass of  $Al_{2}(SO_{4})_{3}$  obtained from $H_{2}SO_{4}$  is less than that from $Al(OH)_{3}$, $H_{2}SO_{4}$ is the limiting reactant.

(b) Number of moles of $H_{2}SO_{4}$ is 0.410 mol, from the balanced chemical reaction 3 moles of $H_{2}SO_{4}$ gives 1 mole of $Al_{2}(SO_{4})_{3}$ thus, 1 mole of $H_{2}SO_{4}$ gives 0.33 moles of $Al_{2}(SO_{4})_{3}$ and 0.410 moles of $H_{2}SO_{4}$  gives 0.136 moles of $Al_{2}(SO_{4})_{3}$.

Therefore, number of moles of $Al_{2}(SO_{4})_{3}$ 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 $H_{2}SO_{4}$ reacts completely to give 0.136 mol of $Al_{2}(SO_{4})_{3}$.

From the balanced chemical reaction, 1 mole of $Al_{2}(SO_{4})_{3}$ formed from 2 moles of $Al(OH)_{3}$, thus, 0.136 mol of $Al_{2}(SO_{4})_{3}$ formed from 0.136×2=0.272 mol.

Thus, 0.272 moles of $Al_{2}(SO_{4})_{3}$ 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.