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Chapter 9: Problem 97

Consider the following unbalanced chemical equation: $$ \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Determine the maximum number of moles of \(\mathrm{SO}_{2}\) produced from \(8.0 \mathrm{moles}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and 3.0 moles of \(\mathrm{O}_{2}\)

Short Answer

Expert verified
The maximum number of moles of \(\mathrm{SO}_{2}\) produced from 8.0 moles of \(\mathrm{H}_{2}\mathrm{S}\) and 3.0 moles of \(\mathrm{O}_{2}\) is 1.5 moles, as \(\mathrm{O}_{2}\) is the limiting reactant.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to make sure that the number of atoms of each element is the same on both sides of the reaction arrow. In this case, we can start by balancing the oxygen atoms. We see that we need 2 moles of \(\mathrm{O}_{2}\) on the reactant side for each mole of the produced \(\mathrm{SO}_{2}\). Also, we need to have 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) to balance the hydrogen atoms in the \(\mathrm{H}_{2} \mathrm{S}\). The balanced chemical equation is: $$ \mathrm{H}_{2} \mathrm{S}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$
02

Identify the limiting reactant

The limiting reactant is the reactant that runs out first and therefore determines the amount of product that can be formed. In this problem, we have 8.0 moles of \(\mathrm{H}_{2} \mathrm{S}\) and 3.0 moles of \(\mathrm{O}_{2}\) given. To find out which reactant is limiting, we compare the given amount of moles of each reactant to the stoichiometric coefficients of the balanced equation: $$\frac{8.0 \ \text{moles}\ \mathrm{H}_{2} \mathrm{S}}{1} = 8.0 \ \text{moles}$$ $$\frac{3.0 \ \text{moles}\ \mathrm{O}_{2}}{2} = 1.5 \ \text{moles} $$ As 1.5 moles is less than 8.0 moles, \(\mathrm{O}_{2}\) is the limiting reactant.
03

Determine the maximum moles of \(\mathrm{SO}_{2}\) produced

Now that we know that oxygen is the limiting reactant and we have 3.0 moles of it, we can calculate the maximum moles of \(\mathrm{SO}_{2}\) produced. Using stoichiometry, we'll convert moles of limiting reactant to moles of product as follow: $$3.0 \ \text{moles} \ \mathrm{O}_{2} \times \frac{1 \ \text{mole} \ \mathrm{SO}_{2}}{2 \ \text{moles} \ \mathrm{O}_{2}} = 1.5 \ \text{moles} \ \mathrm{SO}_{2}$$ Therefore, the maximum number of moles of \(\mathrm{SO}_{2}\) produced is 1.5 moles.

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Key Concepts

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

Limiting Reactant
In any chemical reaction, the limiting reactant, or reagent, is the substance that is entirely consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, because, once it's used up, the reaction cannot continue. Hence, it determines the maximum amount of product that can be formed from the reactants.

To identify the limiting reactant, you need to look at the stoichiometry of the balanced chemical equation. Calculate the moles of each reactant based on its stoichiometric coefficient.
  • Divide the number of moles of each reactant by its respective coefficient in the balanced equation.
  • The reactant with the smallest resulting value is the limiting reactant.
In our exercise, we found that \(\mathrm{O}_{2}\) is the limiting reactant, because, based on its stoichiometry, it allows fewer moles of product to be formed compared to \(\mathrm{H}_{2} \mathrm{~S}\).
Balancing Chemical Equations
Balancing chemical equations involves making sure that there are equal numbers of each type of atom on both sides of the equation. This is crucial because matter cannot be created or destroyed, according to the law of conservation of mass.

To balance an equation:
  • List all elements involved in the reaction on both sides of the equation.
  • Count the number of atoms of each element on both sides.
  • Add coefficients in front of compounds as necessary to make the number of atoms of each element equal on both sides.
In our example, we balanced the equation to gain two moles of \(\mathrm{O}_{2}\) and one mole each of \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). This balancing ensures that for every 2 moles of oxygen used, one mole of each product is produced.
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products. They involve the breaking of old bonds and the formation of new ones, changing the chemical substances into different compounds.

In stoichiometry, understanding chemical reactions involves:
  • Knowing the reactants and predicting the possible products.
  • Applying the concept of the limiting reactant to determine quantities of products formed.
  • Balancing the equation to obey the law of conservation of mass.
Our specific chemical reaction, involving \(\mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{O}_{2}\), results in the transformation of these reactants into \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), showcasing the fundamental aspects of chemical processes and stoichiometry.

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Most popular questions from this chapter

An air bag is deployed by utilizing the following reaction (the nitrogen gas produced inflates the air bag): $$ 2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g) $$ If \(10.5 \mathrm{~g}\) of \(\mathrm{NaN}_{3}\) is decomposed, what theoretical mass of sodium should be produced? If only \(2.84 \mathrm{~g}\) of sodium is actually collected, what is the percent yicld?

When elemental copper is placed in a solution of silver nitrate, the following oxidationreduction reaction takes place, forming clemental silver: $$ \mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s) $$ What mass of copper is required to remove all the silver from a silver nitrate solution containing \(1.95 \mathrm{mg}\) of silver nitrate?

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of each product gas if \(1.25 \mathrm{~g}\) of ammonium nitrate reacts.

Natural waters often contain relatively high levels of calcium ion, \(\mathrm{Ca}^{2+},\) and hydrogen carbonate ion (bicarbonate), \(\mathrm{HCO}_{3}^{-}\), from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, \(\mathrm{CaCO}_{3}\), which forms a deposit ("scale") on the interior of boilers, pipes, and other plumbing fixtures. $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a sample of well water contains \(2.0 \times 10^{-3} \mathrm{mg}\) of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per milliliter, what mass of \(\mathrm{CaCO}_{3}\) scale would \(1.0 \mathrm{~mL}\) of this water be capable of depositing?

For each of the following unbalanced reactions, suppose exactly 5.00 moles of each reactant are taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. For cach reaction, solve the problem three ways: i. Set up and use Before-Change-After (BCA) tables. ii. Compare the moles of reactants to see which runs out first. iii. Consider the amounts of products that can be formed by completcly consuming cach reactant. a. \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) b. \(\operatorname{AgNO}_{3}(a q)+\mathbf{A l}(s) \rightarrow \mathbf{A}_{\mathbf{g}}(s)+\mathbf{A l}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)
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