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Chapter 3: Problem 78

Aluminum hydroxide reacts with sulfuric acid as follows: $2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l)\( Which is the limiting reactant when \)0.500 \mathrm{~mol} \mathrm{Al}(\mathrm{OH})_{3}\( and \)0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}$ are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Short Answer

Expert verified
The limiting reactant in this reaction is H₂SO₄. Under these conditions, 0.167 mol of Al₂(SO₄)₃ can form, and there will be 0.167 mol of excess Al(OH)₃ remaining after the reaction is complete.

Step by step solution

01

Identify the balanced chemical equation

The balanced chemical equation for the reaction is given as: \[2 \mathrm{Al}(\mathrm{OH})_{3}(s) + 3 \mathrm{H}_{2} \mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 6 \mathrm{H}_{2} \mathrm{O}(l)\]
02

Determine the limiting reactant

To determine the limiting reactant, it is necessary to compare the mole ratios for the two reactants: Al(OH)₃ and H₂SO₄. We have 0.500 mol of both reactants. According to the balanced equation, we need 2 moles of Al(OH)₃ for every 3 moles of H₂SO₄. Calculate the mole ratio for each reactant: Mole ratio of Al(OH)₃ = moles of Al(OH)₃ / ratio in the balanced equation = (0.500 mol) / 2 = 0.250 Mole ratio of H₂SO₄ = moles of H₂SO₄ / ratio in the balanced equation = (0.500 mol) / 3 = 0.167 The reactant with the smaller mole ratio is the limiting reactant. In this case, H₂SO₄ is the limiting reactant.
03

Calculate the number of moles of Al₂(SO₄)₃ that form

Now, we will use stoichiometry to determine how many moles of Al₂(SO₄)₃ can form based on the limiting reactant, H₂SO₄. Since the mole ratio in the balanced equation is 3 moles of H₂SO₄ to 1 mole of Al₂(SO₄)₃, we can use the number of moles of H₂SO₄ to find the number of moles of the product: Moles of Al₂(SO₄)₃ = moles of H₂SO₄ × (1 mol Al₂(SO₄)₃ / 3 mol H₂SO₄) = 0.500 mol × (1 / 3) = 0.167 mol of Al₂(SO₄)₃
04

Calculate the remaining moles of the excess reactant

After the reaction is complete, there will be some remaining Al(OH)₃ as it is the excess reactant. We can determine the amount of Al(OH)₃ used in the reaction based on the moles of H₂SO₄ and the mole ratio of Al(OH)₃ to H₂SO₄: Moles of Al(OH)₃ used = moles of H₂SO₄ × (2 mol Al(OH)₃ / 3 mol H₂SO₄) = 0.500 mol × (2 / 3) = 0.333 mol of Al(OH)₃ used To find the remaining moles of Al(OH)₃, subtract the moles used from the initial moles: Remaining moles of Al(OH)₃ = initial moles - moles used = 0.500 mol - 0.333 mol = 0.167 mol of Al(OH)₃
05

Summary

The limiting reactant in this reaction is H₂SO₄. Under these conditions, 0.167 mol of Al₂(SO₄)₃ can form, and there will be 0.167 mol of excess Al(OH)₃ remaining after the reaction is complete.

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of $\mathrm{CO}, 0.733 \mathrm{~g}\( of \)\mathrm{CO}_{2},\( and \)0.450 \mathrm{~g}\( of \)\mathrm{H}_{2} \mathrm{O}$ were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

The molecular formula of salicylic acid, a compound commonly found in facial cleanser, is \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\). (a) What is the molar mass of salicylic acid? (b) How many moles of salicylic acid are present in \(0.5 \mathrm{mg}\) of this substance? (c) How many molecules of salicylic acid are in \(0.5 \mathrm{mg}\) of this substance? (d) How many oxygen atoms are present in \(0.5 \mathrm{mg}\) of salicylic acid?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(0.2 \mathrm{~mol} \mathrm{PCl}_{5}\), molecules, $80 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}, 3.0 \times 10^{23}$ CO molecules.

The molecular formula of saccharin, an artificial sweetener, is \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} .(\mathbf{a})\) What is the molar mass of saccharin? (b) How many moles of sachharin are in $2.00 \mathrm{mg}\( of this substance? (c) How many molecules are in \)2.00 \mathrm{mg}\( of this substance? (d) How many C atoms are present in \)2.00 \mathrm{mg}$ of saccharin?

(a) Ibuprofen is a common over-the-counter analgesic with the formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2} .\) How many moles of \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\) are in a 500-mg tablet of ibuprofen? Assume the tablet is composed entirely of ibuprofen. (b) How many molecules of $\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}$ are in this tablet? (c) How many oxygen atoms are in the tablet?
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