Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 169

Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced chemical equation for this reaction is given below: $$ \mathrm{SO}_{2}(g)+\mathrm{NaOH}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Assuming you react 38.3 g sulfur dioxide with 32.8 g sodium hydroxide and assuming that the reaction goes to completion, calculate the mass of each product formed.

Short Answer

Expert verified
The mass of sodium sulfite (Na2SO3) formed is 75.3 g, and the mass of water (H2O) formed is 10.8 g.

Step by step solution

01

Balance the chemical equation

First, balance the chemical equation for this reaction: $$ \mathrm{SO}_{2}(g) + 2 \mathrm{NaOH}(s) \longrightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l) $$ Now, we have the balanced chemical equation.
02

Convert the given masses of reactants to moles

We are given 38.3 g of sulfur dioxide (SO2) and 32.8 g of sodium hydroxide (NaOH). Convert these masses to moles using their molar masses: Molar mass of SO2: \(1 \cdot 32.1 \thinspace g/mol + 2 \cdot 16.0 \thinspace g/mol = 64.1 \thinspace g/mol\) Moles of SO2: \(\frac{38.3 \thinspace g}{64.1 \thinspace g/mol} = 0.597 \thinspace mol\) Molar mass of NaOH: \(1 \cdot 22.99 \thinspace g/mol + 1 \cdot 15.999 \thinspace g/mol + 1 \cdot 1.00784 \thinspace g/mol = 39.997 \thinspace g/mol\) Moles of NaOH: \(\frac{32.8 \thinspace g}{39.997 \thinspace g/mol} = 0.820 \thinspace mol\)
03

Use stoichiometry to find the limiting reactant

Using the balanced equation, we can determine the limiting reactant by comparing the mole ratios of the reactants: Mole ratio of SO2 to NaOH: \(\frac{0.597 \thinspace mol}{0.820 \thinspace mol} = 0.728\) Since the balanced equation requires a 1:2 ratio of SO2 to NaOH, and the actual mole ratio is less than 1, this means that SO2 is the limiting reactant.
04

Calculate the moles of products formed

Using the stoichiometry from the balanced equation: Moles of Na2SO3: \(0.597 \thinspace mol \cdot \frac{1 \thinspace mol \thinspace Na_{2}SO_{3}}{1 \thinspace mol \thinspace SO_{2}} = 0.597 \thinspace mol\) Moles of H2O: \(0.597 \thinspace mol \cdot \frac{1 \thinspace mol \thinspace H_{2}O}{1 \thinspace mol \thinspace SO_{2}} = 0.597 \thinspace mol\).
05

Convert moles of products to mass

Finally, convert the moles of each product formed to mass using their molar masses: Mass of Na2SO3: \(0.597 \thinspace mol \cdot 126.05 \thinspace g/mol = 75.3 \thinspace g\) Mass of H2O: \(0.597 \thinspace mol \cdot 18.015 \thinspace g/mol = 10.8 \thinspace g\) Hence, 75.3 g of sodium sulfite (Na2SO3) and 10.8 g of water (H2O) are formed as products.

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.

Start your free trial

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

A \(2.25-\) g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron((III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\)

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce $1.0 \times 10^{6} \mathrm{kg}\( \)\mathrm{HNO}_{3}\( by the Ostwald process? Assume 100\)\%$ yield in each reaction, and assume that the NO produced in the third step is not recycled.

The sugar sucrose, which is present in many fruits and vegetables, reacts in the presence of certain yeast enzymes to produce ethanol and carbon dioxide gas. Balance the following equation for this reaction of sucrose. $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{CO}_{2}(g) $$

When aluminum metal is heated with an element from Group 6 \(\mathrm{A}\) of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6 \(\mathrm{A}\) element, the product is 18.56\(\%\) Al by mass. What is the formula of the compound?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation
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