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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 is H₂SO₄, forming 0.167 mol Al₂(SO₄)₃ with 0.167 mol Al(OH)₃ remaining.

Step by step solution

01

Identify Mole Ratios from Balanced Equation

The balanced chemical equation is \(2 \mathrm{Al} (\mathrm{OH})_3 + 3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 6 \mathrm{H}_2 \mathrm{O}\). From this equation, the mole ratio between \(\mathrm{Al} (\mathrm{OH})_3\) and \(\mathrm{H}_2 \mathrm{SO}_4\) is 2:3.
02

Calculate Required Moles of Each Reactant

To use the entire amount of \(0.500 \text{ mol} \) \(\mathrm{Al} (\mathrm{OH})_3\), you need \(0.500 \text{ mol} \times \frac{3}{2} = 0.750 \text{ mol} \) of \(\mathrm{H}_2 \mathrm{SO}_4\). To fully react with the given \(0.500 \text{ mol} \) \(\mathrm{H}_2 \mathrm{SO}_4\), you need \(0.500 \text{ mol} \times \frac{2}{3} = 0.333 \text{ mol} \) of \(\mathrm{Al} (\mathrm{OH})_3\).
03

Identify Limiting Reactant

Since we have only \(0.500 \text{ mol}\) of both \(\mathrm{Al} (\mathrm{OH})_3\) and \(\mathrm{H}_2 \mathrm{SO}_4\), and more \(\mathrm{H}_2 \mathrm{SO}_4\) is needed than is available (0.500 mol available vs 0.750 mol needed), \(\mathrm{H}_2 \mathrm{SO}_4\) is the limiting reactant.
04

Determine Moles of Product Formed

Based on the limiting reactant (\(0.500 \text{ mol} \) \(\mathrm{H}_2 \mathrm{SO}_4\)), we use the reaction stoichiometry. From the balanced equation, 3 moles of \(\mathrm{H}_2 \mathrm{SO}_4\) produce 1 mole of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\). Thus, \(\frac{0.500}{3} = 0.167 \text{ mol} \) of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) can be formed.
05

Calculate Excess Reactant Remaining

Use up 0.333 moles of \(\mathrm{Al} (\mathrm{OH})_3\) (needed to react with 0.500 moles \(\mathrm{H}_2 \mathrm{SO}_4\)), leaving \(0.500 - 0.333 = 0.167 \text{ mol} \) of \(\mathrm{Al} (\mathrm{OH})_3\) unreacted.

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Key Concepts

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

Mole Ratio
In a chemical reaction, the mole ratio signifies the proportion of moles between different reactants and products as determined by a balanced chemical equation. This is essential for converting between amounts of reactants and products. When analyzing the reaction between aluminum hydroxide and sulfuric acid, the equation tells us:
  • 2 moles of \( \mathrm{Al} (\mathrm{OH})_3 \)
  • react with 3 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \)
  • producing 1 mole of \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \)
This 2:3:1 mole ratio is a quantitative measurement that guides how much of each substance is needed or produced. It is essential to calculate the correct amounts to avoid having too little or too much of a reactant when performing reactions. This step is crucial before moving on to determine which reactant will limit product formation.
Stoichiometry
Stoichiometry involves using mole ratios from balanced chemical equations to determine relationships between reactants and products. It is the central framework for understanding and performing calculations in chemistry. In this problem, stoichiometry is used to calculate how many moles of each reactant are necessary for the reaction to occur without leaving any reactant in excess.
For example, if 0.500 moles of \( \mathrm{Al} (\mathrm{OH})_3 \) are present, stoichiometry tells us that 0.750 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \) are needed to fully react with it. Conversely, utilizing 0.500 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \) requires only 0.333 moles of \( \mathrm{Al} (\mathrm{OH})_3 \) to be used completely. This way of calculating helps determine which reactant is the limiting one and therefore limits the amount of product that can be formed.
Excess Reactant
In any chemical reaction, the reactant that remains after the complete consumption of the limiting reactant is termed the excess reactant. It is vital to know how much of this reactant remains because it affects the reaction's efficiency and cost.
Using our reaction, we find that \( \mathrm{H}_2 \mathrm{SO}_4 \) is the limiting reactant, leaving \( \mathrm{Al} (\mathrm{OH})_3 \) as the excess reactant. With 0.500 moles of \( \mathrm{Al} (\mathrm{OH})_3 \), only 0.333 moles are required when \( \mathrm{H}_2 \mathrm{SO}_4 \) runs out. Thus, 0.167 moles remain unreacted.

This calculation is necessary for understanding the actual consumption of materials in a reaction and optimizing for future reactions.
Chemical Reactions
A chemical reaction involves the transformation of reactants into products through the breaking and forming of bonds. It is a process fundamental to creating new materials and substances. The given reaction is balanced, meaning that the number of atoms for each element is conserved across the reactants and products. Aluminum hydroxide \( (\mathrm{Al} (\mathrm{OH})_3) \) reacts with sulfuric acid \( (\mathrm{H}_2 \mathrm{SO}_4) \), resulting in the formation of aluminum sulfate \( (\mathrm{Al}_2(\mathrm{SO}_4)_3) \) and water \( (\mathrm{H}_2 \mathrm{O}) \).
Balanced chemical equations are crucial for determining the proportions in which chemicals react and the products yield. Not only do they show what substances are consumed and formed, but they also provide the necessary groundwork for the stoichiometric and limiting reactant calculations discussed earlier.

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Most popular questions from this chapter

Detonation of nitroglycerin proceeds as follows: $$ \begin{aligned} 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l) & \longrightarrow \\ & 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) If a sample containing \(2.00 \mathrm{~mL}\) of nitroglycerin (density = \(1.592 \mathrm{~g} / \mathrm{mL})\) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies \(55 \mathrm{~L}\) under the conditions of the explosion, how many liters of gas are produced? (c) How many grams of \(\mathrm{N}_{2}\) are produced in the detonation?

Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains \(75.69 \% \mathrm{C}\) \(8.80 \% \mathrm{H},\) and \(15.51 \% \mathrm{O}\) by mass and has a molar mass of \(206 \mathrm{~g} / \mathrm{mol}\). (b) Cadaverine, a foul-smelling substance produced by the action of bacteria on meat, contains \(58.55 \% \mathrm{C}\), \(13.81 \% \mathrm{H},\) and \(27.40 \% \mathrm{~N}\) by mass; its molar mass is \(102.2 \mathrm{~g} / \mathrm{mol}\) (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains \(59.0 \%\) C, \(7.1 \%\) H, \(26.2 \%\) O, and \(7.7 \%\) N by mass; its molar mass is about \(180 \mathrm{u}\).

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\). (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

Automotive air bags inflate when sodium azide, \(\mathrm{NaN}_{3}\), rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of 1.50 mol of \(\mathrm{NaN}_{3}\) ? (b) How many grams of \(\mathrm{NaN}_{3}\) are required to form \(10.0 \mathrm{~g}\) of nitrogen gas? (c) How many grams of \(\mathrm{NaN}_{3}\) are required to produce \(10.0 \mathrm{ft}^{3}\) of nitrogen gas, about the size of an automotive air bag, if the gas has a density of \(1.25 \mathrm{~g} / \mathrm{L}\) ?

Very small semiconductor crystals, composed of approximately 1000 to 10,000 atoms, are called quantum dots. Quantum dots made of the semiconductor CdSe are now being used in electronic reader and tablet displays because they emit light efficiently and in multiple colors, depending on dot size. The density of CdSe is \(5.82 \mathrm{~g} / \mathrm{cm}^{3}\). (a) What is the mass of one \(2.5-\mathrm{nm}\) CdSe quantum dot? (b) CdSe quantum dots that are \(2.5 \mathrm{nm}\) in diameter emit blue light upon stimulation. Assuming that the dot is a perfect sphere and that the empty space in the dot can be neglected, calculate how many Cd atoms are in one quantum dot of this size. (c) What is the mass of one \(6.5-\mathrm{nm}\) CdSe quantum dot? (d) CdSe quantum dots that are \(6.5 \mathrm{nm}\) in diameter emit red light upon stimulation. Assuming that the dot is a perfect sphere, calculate how many Cd atoms are in one quantum dot of this size. (e) If you wanted to make one \(6.5-\mathrm{nm}\) dot from multiple \(2.5-\mathrm{nm}\) dots, how many \(2.5-\mathrm{nm}\) dots would you need, and how many CdSe formula units would be left over, if any?

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