\(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) 4.0 \(\mathrm{mol}\) of gaseous \(\mathrm{SO}_{2}\) and 6.0 mol of \(\mathrm{O}_{2}\) gas are allowed to react in a sealed container.

Short Answer

Expert verified
4.0 mol of SO3 is produced and 4.0 mol of O2 remains unreacted.

Step by step solution

01

Understand the Molar Ratios

From the balanced chemical equation \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\), we know that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
02

Calculate how many moles of O2 are required for 4.0 mol of SO2

Based on the equation, 2 moles of SO2 need 1 mole of O2. So, for 4.0 mol of SO2, we need 4.0/2 = 2.0 mol of O2.
03

Determine Limiting Reactant

Since we have 6.0 mol of O2, there is more than enough O2 to react with all available SO2. Therefore, SO2 is the limiting reactant.
04

Calculate moles of SO3 produced

Since the reaction ratio of SO2:SO3 is 2:2 (which simplifies to 1:1), the moles of SO3 produced will equal to the moles of SO2 used, i.e., 4.0 mol.
05

Calculate remaining O2

The moles of O2 remaining will be initial moles of O2 - moles of O2 used = 6.0 - 2.0 = 4.0 mol.

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!

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

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. What is the approximate amount of heat released during the reaction? \(\begin{array}{ll}{\text { (A) }} & {1.5 \mathrm{kJ}} \\ {\text { (B) }} & {2.9 \mathrm{kJ}} \\ {\text { (C) }} & {5.9 \mathrm{kJ}} \\ {\text { (D) }} & {11.8 \mathrm{kJ}}\end{array}\)

Nitrogen’s electronegativity value is between those of phosphorus and oxygen. Which of the following correctly describes the relationship between the three values? (A) The value for nitrogen is less than that of phosphorus because nitrogen is larger, but greater than that of oxygen because nitrogen has a greater effective nuclear charge. (B) The value for nitrogen is less than that of phosphorus because nitrogen has fewer protons, but greater than that of oxygen because nitrogen has fewer valence electrons. (C) The value for nitrogen is greater than that of phosphorus because nitrogen has fewer electrons, but less than that of oxygen because nitrogen is smaller. (D) The value for nitrogen is greater than that of phosphorus because nitrogen is smaller, but less than that of oxygen because nitrogen has a smaller effective nuclear charge.

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. During this reaction, heat transfers from (A) the reactants to the products (B) the reactants to the system (C) the system to the surroundings (D) the products to the surroundings

\(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{PCl}_{5}(g) \Delta H=-92.5 \mathrm{kJ} / \mathrm{mol}\) In which of the following ways could the reaction above be manipulated to create more product? (A) Decreasing the concentration of \(\mathrm{PCl}_{3}\) (B) Increasing the pressure (C) Increasing the temperature (D) None of the above

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