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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

When elemental carbon is burned in the open atmosphere, with plenty of oxygen gas prescnt, the product is carbon dioxide. $$ \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) $$ However, when the amount of oxygen present during the burning of the carbon is restricted, carbon monoxide is more likely to result. $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}(g) $$ What mass of each product is expected when a \(5.00-\mathrm{g}\) sample of pure carbon is burned under cach of these conditions?

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. \(\operatorname{AgNO}_{3}(a q)+\mathbf{A l}(s) \rightarrow \mathbf{A}_{\mathbf{g}}(s)+\mathbf{A l}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

For each of the following unbalanced equations, indicate how many moles of the second reactant would be required to react exactly with 0.275 mole of the first reactant. State clearly the mole ratio used for the conversion. a. \(\mathrm{Cl}_{2}(g)+\mathrm{KI}(a q) \rightarrow \mathrm{I}_{2}(s)+\mathrm{KCl}(a q)\) b. \(\mathrm{Co}(s)+\mathrm{P}_{4}(s) \rightarrow \mathrm{Co}_{3} \mathrm{P}_{2}(s)\) \(\mathrm{c} \cdot \mathrm{Zn}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{ZnNO}_{3}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{C}_{3} \mathrm{H}_{12}(t)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

One process for the commercial production of baking soda (sodium hydrogen carbonate) involves the following reaction, in which the carbon dioxide is used in its solid form ("dry ice" enough for the sodium hydrogen carbonate to precipitate: $$ \mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(s) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaHCO}_{3}(s) $$ Because they are relatively cheap, sodium chloride and water are typically present in excess. What is the expected yield of \(\mathrm{NaHCO}_{3}\) when one performs such a synthesis using \(10.0 \mathrm{~g}\) of ammonia and \(15.0 \mathrm{~g}\) of dry ice, with an excess of \(\mathrm{NaCl}\) and water?

Although we tend to make less use of mercury these days because of the cnvironmental problems created by its improper disposal, mercury is still an important metal because of its unusual property of existing as a liquid at room temperature. One process by which mercury is produced industrially is through the heating of its common ore cinnabar (mercuric sulfide, \(\mathrm{HgS}\) ) with lime (calcium oxide, \(\mathrm{CaO}\) ). $$ 4 \mathrm{HgS}(s)+4 \mathrm{CaO}(s) \rightarrow 4 \mathrm{Hg}(l)+3 \mathrm{CaS}(s)+\mathrm{CaSO}_{4}(s) $$ What mass of mercury would be produced by complete reaction of \(10.0 \mathrm{~kg}\) of \(\mathrm{HgS}\) ?

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