For the reaction shown, calculate the theoretical yield of the product in moles for each of the initial quantities of reactants. $$ \mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s) $$ (a) \(2 \mathrm{~mol} \mathrm{Ti} ; 2 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(5 \mathrm{~mol} \mathrm{Ti} ; 9 \mathrm{~mol} \mathrm{Cl}_{2}\) (c) \(0.483 \mathrm{~mol} \mathrm{Ti} ; 0.911 \mathrm{~mol} \mathrm{Cl}_{2}\) (d) \(12.4 \mathrm{~mol} \mathrm{Ti} ; 15.8 \mathrm{~mol} \mathrm{Cl}_{2}\)

Short Answer

Expert verified
The theoretical yields of TiCl4 are: (a) 2 moles, (b) 4.5 moles, (c) 0.4555 moles, (d) 7.9 moles.

Step by step solution

01

Write the balanced chemical equation

Confirm that the chemical equation given is balanced. It is already given in the balanced form: \(\mathrm{Ti}(s) + 2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s)\). This indicates that 1 mole of Ti reacts with 2 moles of Cl2 to form 1 mole of TiCl4.
02

Identify the limiting reactant for each scenario

The limiting reactant is the one that will be completely consumed first and thus determines the amount of product formed. For each reaction scenario (a)-(d), compare the mole ratio of reactants to the coefficients in the balanced equation to find the limiting reactant.
03

Calculate the theoretical yield for each reaction scenario

Using the stoichiometry of the balanced equation and the limiting reactant determined in Step 2, calculate the theoretical yield in moles of TiCl4 for each scenario.
04

Scenario (a)

For (a) we have equal moles of Ti and Cl2, but since the equation requires two moles of Cl2 for every mole of Ti, Cl2 is the limiting reactant. The theoretical yield of TiCl4 is the same as the moles of Ti present, which is 2 moles.
05

Scenario (b)

For (b) we have 5 moles of Ti and 9 moles of Cl2. This is not enough chlorine to react with all the titanium. The limiting reactant is Cl2, which will be consumed after 4.5 moles of Ti have reacted. Thus, the theoretical yield of TiCl4 is 4.5 moles.
06

Scenario (c)

For (c) we have 0.483 moles of Ti and 0.911 moles of Cl2. The chlorine is again the limiting reactant, and all of it will react with half as much Ti to give 0.4555 moles of TiCl4 as a theoretical yield.
07

Scenario (d)

For (d) we have 12.4 moles of Ti and 15.8 moles of Cl2. The limiting reactant is again Cl2, which will react with 7.9 moles of Ti. Therefore, the theoretical yield of TiCl4 is 7.9 moles.

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Key Concepts

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

Stoichiometry
Stoichiometry is the section of chemistry that involves quantitative relationships between the reactants and products in a chemical reaction. Think of stoichiometry as the recipe for a chemical reaction, where the balanced chemical equation provides the quantities of each reactant that reacts and the amount of product that will form. Just like baking a cake requires specific amounts of each ingredient, a chemical reaction requires specific ratios of reactants to produce products.

To perform stoichiometric calculations, you need to start with a balanced chemical equation. This equation tells you the molar ratio between reactants and products. For instance, in the given exercise, the balanced chemical equation is \(\mathrm{Ti}(s) + 2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s)\), which signifies 1 mole of titanium (Ti) reacts with 2 moles of chlorine gas (\(\mathrm{Cl}_{2}\)) to produce 1 mole of titanium tetrachloride (\(\mathrm{TiCl}_{4}\)).

Using this ratio, we can calculate how much product can be formed from given amounts of reactants. This is an essential tool in chemistry because it helps predict the quantities needed for a reaction to take place fully or to find out how much product we can expect to produce.
Limiting Reactant
The concept of the limiting reactant is crucial in predicting the amount of product formed in a chemical reaction. The limiting reactant is the substance that is totally consumed when the chemical reaction is complete. Once this reactant is used up, the reaction stops, and no more product can be formed, even if other reactants are still available. It's the bottleneck of the reaction, limiting the amount of product that can be generated.

In the exercise provided, we determine the limiting reactant by comparing the mole ratio of the reactants to the ratios in the balanced equation. For instance, in scenario (a), we have 2 moles of titanium and 2 moles of chlorine gas. However, the titanium requires twice as much chlorine gas to react completely, making chlorine gas the limiting reactant, as only 1 mole of titanium can react with the 2 moles of chlorine gas present.

Identifying the limiting reactant is an essential step in stoichiometry. It not only determines the theoretical yield of a reaction but can also be utilized in industry to minimize waste and optimize resource usage. Students often struggle with this concept, so it's important to practice with various examples to become comfortable in finding the limiting reactant in chemical reactions.
Balanced Chemical Equations
Balanced chemical equations are foundational in chemistry—they represent the principle of conservation of mass, stating that matter is neither created nor destroyed in a typical chemical reaction. A balanced equation must have the same numbers of each type of atom on both sides of the reaction arrow.

The balanced chemical equation provided in the exercise, \(\mathrm{Ti}(s) + 2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s)\), shows a simple 1:2:1 ratio among titanium, chlorine gas, and titanium tetrachloride respectively. This tells us how many moles of each reactant are needed to produce a certain amount of product, and vice versa. To solve stoichiometric problems, the first critical step is always to write and confirm the balanced chemical equation. This step sets the stage for all subsequent calculations, such as identifying the limiting reactant and calculating theoretical yield.

Understanding how to balance chemical equations is an essential skill for any student studying chemistry. It not only is crucial for laboratory work but also helps to reinforce the understanding of atomic structure and conservation laws in chemical reactions.

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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. $$ 2 \mathrm{~K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{KCl}(s) $$ (a) \(1 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(1.8 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (c) \(2.2 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (d) \(14.6 \mathrm{~mol} \mathrm{~K} ; 7.8 \mathrm{~mol} \mathrm{Cl}_{2}\)

Consider the unbalanced equation for the combustion of butane: $$ \mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Balance the equation and determine how many moles of \(\mathrm{O}_{2}\) are required to react completely with \(4.9 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10}\).

For each reaction, calculate how many moles of product form when \(1.75 \mathrm{~mol}\) of the reactant in color completely reacts. Assume there is more than enough of the other reactant. (a) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(2 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (d) \(2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\)

Consider the reaction between \(\mathrm{N}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) : $$ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ A reaction vessel initially contains \(27.5 \mathrm{~g} \mathrm{~N}_{2} \mathrm{H}_{4}\) and \(74.9 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\). Calculate the masses of \(\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) that will be in the reaction vessel after the reactants have reacted as much as possible. Assume \(100 \%\) yield. Hint: The limiting reactant is completely consumed, but the reactant in excess is not. Use the amount of limiting reactant to determine the amount of products that form and the amount of the reactant in excess that remains after complete reaction.

Sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right)\) reacts with water to form sodium hydroxide and oxygen gas. Write a balanced equation for the reaction and determine how much oxygen in grams is formed by the complete reaction of \(35.23 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\).

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