Consider the reaction between sulfur trioxide and water: $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ A chemist allows \(61.5 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(11.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) to react. When the reaction is finished, the chemist collects \(54.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Short Answer

Expert verified
The limiting reactant is water (H2O), the theoretical yield is calculated from the moles of water, and the percent yield is calculated by dividing the actual yield (54.9 g of H2SO4) by the theoretical yield and multiplying by 100.

Step by step solution

01

- Calculate the number of moles of each reactant

Use the molar mass of each reactant to convert the mass of each reactant to moles. For sulfur trioxide \( \mathrm{SO}_{3} \) its molar mass is approximately 80.07 g/mol. For water \( \mathrm{H}_{2}O \) its molar mass is approximately 18.02 g/mol. Moles of \( \mathrm{SO}_{3} \) are calculated by \( \frac{61.5 \text{ g}}{80.07 \text{ g/mol}} \) and moles of \( \mathrm{H}_{2}O \) are calculated by \( \frac{11.2 \text{ g}}{18.02 \text{ g/mol}} \) .
02

- Determine the limiting reactant

Using the stoichiometry of the reaction which is 1:1, compare the molar amounts of the reactants. The reactant that yields the least amount of product is the limiting reactant.
03

- Calculate the theoretical yield

The theoretical yield is the amount of product formed when the limiting reactant is completely consumed. It can be found by using the stoichiometry of the reaction and the moles of the limiting reactant.
04

- Find the percent yield

Percent yield is calculated using the formula \(\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\text{%}\). The actual yield is given as 54.9 g of \( \mathrm{H}_{2}\mathrm{SO}_{4} \). The theoretical yield obtained from the previous step will be used to calculate the percent yield.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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 amounts of reactants and products in a chemical reaction. It is based on the conservation of mass and the concept that atoms are neither created nor destroyed during a chemical reaction.

To perform stoichiometric calculations, it is essential to understand and use the mole concept, which provides a way to count atoms, ions, and molecules in bulk amounts by weighing them. Once the number of moles is known for each reactant, their ratio can be compared to the stoichiometric ratio given by the balanced chemical equation. This allows us to determine which reactant will run out first and thus is the limiting reactant.
Percent Yield
The percent yield is a measure of the efficiency of a chemical reaction. To find the percent yield, we compare the actual yield—the amount of product actually obtained from the reaction—to the theoretical yield—the calculated amount of product that should have been obtained if the reaction had gone perfectly with 100% efficiency.

The formula for percent yield is:
\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]
This calculation is a crucial step in determining the success of a chemical synthesis and enables chemists to identify and improve reaction conditions.
Theoretical Yield
The theoretical yield is the predicted amount of product generated in a chemical reaction, assuming it goes to completion with no side reactions and 100% efficiency. To calculate the theoretical yield, one must consider the balanced chemical equation and the limiting reactant, which is the reactant that will be completely consumed first.

The theoretical yield serves as a reference to which the actual yield can be compared. It is a key indicator for evaluating the percentage of a reaction's successful conversion of reactants into the desired product. Understanding how to accurately calculate this yield is fundamental in both academic and industrial chemical labs.
Molar Mass
Molar mass is a physical property of substances that is crucial for stoichiometric calculations in chemistry. It is defined as the mass in grams of one mole of the substance. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\).

The molar mass of a compound can be determined by summing the atomic masses of each element present in one molecule of the compound, as found on the periodic table. For instance, to find the molar mass of \(\mathrm{SO}_{3}\), one would multiply the atomic mass of sulfur by one (since there's one sulfur atom) and the atomic mass of oxygen by three (since there are three oxygen atoms), and then add the results together. Accurate calculation of molar mass is essential for converting between mass and moles, which is the foundational step in performing stoichiometric computations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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) \(2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\)

For the reaction shown, calculate how many moles of \(\mathrm{NH}_{3}\) form when each amount of reactant completely reacts. $$ 3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g) $$ (a) \(5.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (b) \(2.28 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (c) \(5.8 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (d) \(9.76 \times 10^{7} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\)

Determine the theoretical yield of \(C\) when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react in the generic reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ (a) \(1 \mathrm{~mol} \mathrm{~A} ; 1 \mathrm{~mol} \mathrm{~B}\) (b) \(2 \mathrm{~mol} \mathrm{~A} ; 2 \mathrm{~mol} \mathrm{~B}\) (c) \(1 \mathrm{~mol} \mathrm{~A} ; 3 \mathrm{~mol} \mathrm{~B}\) (d) \(32 \mathrm{~mol} \mathrm{~A} ; 68 \mathrm{~mol} \mathrm{~B}\)

For each reaction, calculate how many moles of the product form when \(0.112 \mathrm{~mol}\) of the reactant in color completely reacts. Assume there is more than enough of the other reactant. (a) \(2 \mathrm{Ca}(\mathrm{s})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CaO}(s)\) (b) \(4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(4 \mathrm{~K}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~K}_{2} \mathrm{O}(s)\) (d) \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\)

Consider the recipe for making tomato and garlic pasta. 2 cups noodles \(+12\) tomatoes \(+3\) cloves garlic \(\longrightarrow\) 4 servings pasta If you have 7 cups of noodles, 27 tomatoes, and 9 cloves of garlic, how many servings of pasta can you make? Which ingredient limits the amount of pasta that it is possible to make?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free