Consider the unbalanced equation for the reaction of aluminum with sulfuric acid: $$ \mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g) $$ (a) Balance the equation. (b) How many moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are required to completely react with \(8.3 \mathrm{~mol}\) of \(\mathrm{Al}\) ? (c) How many moles of \(\mathrm{H}_{2}\) are formed by the complete reaction of \(0.341 \mathrm{~mol}\) of \(\mathrm{Al}\) ?

Short Answer

Expert verified
The balanced equation is \(2\mathrm{Al}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 3\mathrm{H}_{2}(g)\). (b) 12.45 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) are required to react with 8.3 moles of \(\mathrm{Al}\). (c) 0.5115 moles of \(\mathrm{H}_{2}\) are formed by the complete reaction of 0.341 moles of \(\mathrm{Al}\).

Step by step solution

01

Balance Aluminum (Al)

First, balance the aluminum atoms. There are 2 aluminum atoms on the product side in \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), so we need 2 aluminum atoms on the reactant side. The balanced equation for aluminum is now \(2\mathrm{Al}(s) + \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + \mathrm{H}_{2}(g)\).
02

Balance Sulfur

There are 3 sulfate ions \(\mathrm{SO}_{4}^{2-}\) on the product side, which means there are 3 sulfur atoms that need to be balanced on the reactant side. This requires 3 molecules of sulfuric acid. The equation now looks like \(2\mathrm{Al}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + \mathrm{H}_{2}(g)\).
03

Balance Hydrogen and Oxygen

There are a total of 6 hydrogen atoms on the reactant side now. To balance the hydrogen atoms on the product side, we place a coefficient of 3 in front of \(\mathrm{H}_{2}(g)\). The oxygens are already balanced as they are part of the sulfate group. The fully balanced equation is \(2\mathrm{Al}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 3\mathrm{H}_{2}(g)\).
04

Calculate Moles of H2SO4 Required for 8.3 mol Al

Using the balanced equation, for every 2 moles of \(\mathrm{Al}\), we need 3 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\). Thus, for 8.3 mol of \(\mathrm{Al}\), we calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) as: \[(8.3 \, \mathrm{mol} \, \mathrm{Al}) \times \frac{3 \, \mathrm{mol} \, \mathrm{H}_{2}\mathrm{SO}_{4}}{2 \, \mathrm{mol} \, \mathrm{Al}} = 12.45 \, \mathrm{mol} \, \mathrm{H}_{2}\mathrm{SO}_{4}\].
05

Calculate Moles of H2 Formed from 0.341 mol Al

Using the balanced equation, 2 moles of \(\mathrm{Al}\) produce 3 moles of \(\mathrm{H}_{2}\). For 0.341 mol of \(\mathrm{Al}\), we calculate the moles of \(\mathrm{H}_{2}\) as: \[(0.341 \, \mathrm{mol} \, \mathrm{Al}) \times \frac{3 \, \mathrm{mol} \, \mathrm{H}_{2}}{2 \, \mathrm{mol} \, \mathrm{Al}} = 0.5115 \, \mathrm{mol} \, \mathrm{H}_{2}\].

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Understanding stoichiometry is crucial because it allows chemists to predict the amounts of substances consumed and produced in a reaction.

In any balanced chemical equation, the coefficients represent the stoichiometric ratios, which tell us how many moles of each reactant and product are involved in the reaction. For instance, in the balanced equation of the reaction between aluminum and sulfuric acid, the stoichiometric ratios are essential to determine how much sulfuric acid is required to react with a given amount of aluminum, or how much hydrogen gas is produced in the process.

Keep in mind the importance of the mole concept and molar ratios in stoichiometry. These ratios serve as a conversion factor between the number of moles of different substances in a reaction, something you will encounter in virtually every stoichiometric calculation. It’s vital to start with a balanced chemical equation, as any stoichiometric calculation relies on the proportional relationships that the balanced equation provides.
Chemical Reaction
A chemical reaction is a process where substances, known as reactants, transform into new substances called products. During a chemical reaction, the molecular structures of the reactants undergo a change, which results in different physical and chemical properties in the products.

Our example involves a reaction between aluminum and sulfuric acid. Initially, the atoms of aluminum and the compound sulfuric acid are rearranged to form aluminum sulfate and hydrogen gas. This transformation is represented symbolically through a chemical equation. Balancing this equation is crucial as it ensures that the law of conservation of mass is upheld, meaning the number of atoms of each element remains constant before and after the reaction.

To fully understand how a chemical reaction occurs, one must consider not only the substances involved but also their physical states, signified by symbols like (s) for solids, (l) for liquids, (g) for gases, and (aq) for aqueous solutions. These details are crucial for accurately conveying the conditions under which a reaction takes place.
Mole Concept
The mole concept is a fundamental aspect of chemistry that facilitates the counting of particles, such as atoms, molecules, or ions, in a given sample. A mole is defined as the amount of a substance that contains as many particles as there are atoms in exactly 12 grams of carbon-12, which is approximately 6.022 x 1023 particles, a number known as Avogadro’s number.

When dealing with chemical reactions, the mole concept allows chemists to measure quantities of reactants and products in a convenient way. In our example, knowing the moles of aluminum lets us determine the moles of sulfuric acid needed or the moles of hydrogen gas produced using the stoichiometric ratios from the balanced chemical equation.

Often times, one of the challenges students face in using the mole concept is the conversion between grams and moles. Remember that to convert from mass to moles, the molar mass of the substance (mass of one mole of the substance) is needed, which can be found on the periodic table or calculated from it. Similarly, to convert from moles to particles, Avogadro’s number is used. These conversions are an integral part of solving stoichiometric problems.

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Most popular questions from this chapter

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ (a) \(1.0 \mathrm{~g} \mathrm{Al} ; 1.0 \mathrm{~g} \mathrm{O}_{2}\) (b) \(2.2 \mathrm{~g} \mathrm{Al} ; 1.8 \mathrm{~g} \mathrm{O}_{2}\) (c) \(0.353 \mathrm{~g} \mathrm{Al} ; 0.482 \mathrm{~g} \mathrm{O}_{2}\)

Consider the generic chemical reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C} $$ How many moles of \(C\) are formed upon complete reaction of: (a) \(2 \mathrm{~mol}\) of \(\mathrm{A}\) (b) \(2 \mathrm{~mol}\) of \(\mathrm{B}\) (c) \(3 \mathrm{~mol}\) of \(\mathrm{A}\) (d) \(3 \mathrm{~mol}\) of \(\mathrm{B}\)

Consider the generic reaction between reactants \(\mathrm{A}\) and \(\mathrm{B}\) : $$ 3 \mathrm{~A}+4 \mathrm{~B} \longrightarrow 2 \mathrm{C} $$ If a reaction vessel initially contains \(9 \mathrm{~mol} \mathrm{~A}\) and \(8 \mathrm{~mol} \mathrm{~B}\), how many moles of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) will be in the reaction vessel after the reactants have reacted as much as possible? (Assume \(100 \%\) actual yield.)

Consider the unbalanced equation for the reaction of solid lead with silver nitrate: $$ \mathrm{Pb}(s)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ag}(s) $$ (a) Balance the equation. (b) How many moles of silver nitrate are required to completely react with \(9.3 \mathrm{~mol}\) of lead? (c) How many moles of Ag are formed by the complete reaction of \(28.4\) mol of \(\mathrm{Pb}\) ?

Consider the unbalanced equation for the neutralization of acetic acid: $$ \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q) $$ Balance the equation and determine how many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to completely neutralize \(1.07 \mathrm{~mol}\) of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\).

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