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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation plays a crucial role in describing a chemical reaction. It shows the reactants transforming into the products. At its core, balancing ensures that atoms are conserved throughout the reaction, meaning the number of each type of atom remains the same on both sides of the equation. This is in line with the law of conservation of mass, which dictates that mass can neither be created nor destroyed.Consider the equation for the reaction between aluminum hydroxide and sulfuric acid: \[2 \mathrm{Al}(\mathrm{OH})_{3}(s) + 3 \mathrm{H}_{2} \mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 6 \mathrm{H}_{2} \mathrm{O}(l)\]The balanced equation tells us that:
  • 2 moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) react with 3 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\).
  • These reactants produce 1 mole of \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) and 6 moles of water \(\mathrm{H}_{2}\mathrm{O}\).
Balancing provides a clear map of how substances interact during the reaction. By understanding this equation, you can dive deeper into other chemical concepts like stoichiometry and limiting and excess reactants.
Stoichiometry
Stoichiometry is the bridge between a balanced chemical equation and the real-world quantities of reactants and products involved in a chemical reaction. It allows us to calculate the amount of substances that participate in and result from reactions, using the mole ratios present in the balanced equation.For our reaction between aluminum hydroxide and sulfuric acid:
  • The mole ratio of \(\mathrm{Al}(\mathrm{OH})_{3}\) to \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) is 2:1.
  • The mole ratio of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) to \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) is 3:1.
Stoichiometry uses these ratios to determine how many moles of product can be formed from a given amount of reactants. For instance, if you start with 0.500 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), stoichiometry helps you calculate that you can produce 0.167 moles of \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), by the relation \(0.500 \text{ mol} \times (1 \text{ mol } \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} / 3 \text{ mol } \mathrm{H}_{2}\mathrm{SO}_{4})\). Thus, stoichiometry provides a quantitative method to approach chemical reactions, making it fundamental in laboratory calculations and real-world applications.
Excess Reactant
In any chemical reaction, the reactants are not always present in the exact stoichiometric amounts needed for complete conversion into products. The limiting reactant is the one that is exhausted first, limiting the amount of product that can be formed, while the excess reactant is what remains after the reaction stops.In our example, given 0.500 mol of both aluminum hydroxide and sulfuric acid:
  • Sulfuric acid \((\mathrm{H}_{2}\mathrm{SO}_{4})\) is the limiting reactant because it is used up first based on its mole ratio in the balanced equation.
  • Aluminum hydroxide \((\mathrm{Al}(\mathrm{OH})_{3})\) becomes the excess reactant.
When calculating remaining quantities, we find that after the reaction, 0.167 mol of \((\mathrm{Al}(\mathrm{OH})_{3})\) is left unreacted. It's important to track the excess reactant, as it can affect the purity of the final product or be key in further calculations or reactions.

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Most popular questions from this chapter

The thermite reaction, $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{Fe} $$ produces so much heat that the Fe product melts. This reaction is used industrially to weld metal parts under water, where a torch cannot be employed. It is also a favorite chemical demonstration in the lecture hall (on a small scale). (a) Balance the chemical equation for the thermite reaction, and include the proper states of matter. (b) Calculate how many grams of aluminum are needed to completely react with \(500.0 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in this reaction. (c) This reaction produces \(852 \mathrm{~kJ}\) of heat per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacted. How many grams of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are needed to produce \(1.00 \times 10^{4} \mathrm{~kJ}\) of heat? (d) If you performed the reverse reaction- aluminum oxide plus iron makes iron oxide plus aluminum-would that reaction have heat as a reactant or a product?

Balance the following equations: (a) \(\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)\) (d) \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)\)

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is \(2.1 \mathrm{~g} / \mathrm{cm}^{3}\). It can be made by reacting boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3}\), with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains \(850 \mathrm{~g}\) of Acrilan \(^{\circledast}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A \(0.1005-g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?
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