/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Potassium superoxide, \(\mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Potassium superoxide, \(\mathrm{KO}_{2},\) reacts with carbon dioxide to form potassium carbonate and oxygen:\(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\).This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much \(\mathrm{O}_{2}\) could be produced from \(2.50 \mathrm{g}\) of \(\mathrm{KO}_{2}\) and \(4.50 \mathrm{g}\) of \(\mathrm{CO}_{2} ?\)

Short Answer

Expert verified
Answer: 0.845 grams of oxygen can be produced from 2.50 g of KOâ‚‚ and 4.50 g of COâ‚‚.

Step by step solution

01

1. Calculate molar masses

First, we need to calculate the molar masses of KOâ‚‚ and COâ‚‚. For KOâ‚‚: - Potassium (K): 39.10 g/mol - Oxygen (O): 16.00 g/mol The molar mass of KOâ‚‚ = 39.10 g/mol + 2 * 16.00 g/mol = 71.10 g/mol For COâ‚‚: - Carbon (C): 12.01 g/mol The molar mass of COâ‚‚ = 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
02

2. Convert masses to moles

Now, we'll convert the given masses of reactants to moles using molar masses: moles of KOâ‚‚ = (2.50 g) / (71.10 g/mol) = 0.0352 mol moles of COâ‚‚ = (4.50 g) / (44.01 g/mol) = 0.102 mol
03

3. Determine the limiting reactant

We'll use the stoichiometry of the reaction to find the limiting reactant. The balanced chemical equation is: \(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\) For every 4 moles of KOâ‚‚, 2 moles of COâ‚‚ are required. So we'll compare the mole ratio of both reactants: Mole ratio = moles of KOâ‚‚ / moles of COâ‚‚ = 0.0352 mol / 0.102 mol = 0.344 Since the mole ratio (0.344) is less than the stoichiometric ratio of 4:2 (or 2), KOâ‚‚ is the limiting reactant.
04

4. Calculate the moles of produced oxygen

According to the balanced chemical equation, 4 moles of KOâ‚‚ produce 3 moles of Oâ‚‚. So, we'll use the moles of the limiting reactant (KOâ‚‚) to calculate the moles of produced oxygen: moles of Oâ‚‚ = (3 moles Oâ‚‚ / 4 moles KOâ‚‚) * moles of KOâ‚‚ = (3 Oâ‚‚ / 4 KOâ‚‚) * 0.0352 mol = 0.0264 mol
05

5. Calculate the mass of produced oxygen

Finally, we'll convert the moles of produced oxygen to mass using the molar mass of Oâ‚‚: - Oxygen (O): 16.00 g/mol The molar mass of Oâ‚‚ = 2 * 16.00 g/mol = 32.00 g/mol mass of Oâ‚‚ = moles of Oâ‚‚ * molar mass of Oâ‚‚ = 0.0264 mol * 32.00 g/mol = 0.845 g Thus, 0.845 grams of oxygen could be produced from 2.50 g of KOâ‚‚ and 4.50 g of COâ‚‚.

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 91Ó°ÊÓ!

Key Concepts

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

Molar Mass Calculation
Molar mass is a fundamental concept in chemistry that allows us to convert between mass and moles, which is crucial for understanding chemical reactions. The molar mass is the mass of one mole of a substance. It is typically expressed in grams per mole (g/mol). To find this value, simply sum up the atomic masses of each element in a compound. These atomic masses can be found on the periodic table.
For potassium superoxide ( K O _2 ), we have:
  • Potassium ( K ) has an atomic mass of 39.10 g/mol.
  • Oxygen ( O ) has an atomic mass of 16.00 g/mol.
So, the molar mass of K O _2 is calculated as:\[ ext{Molar mass of } KO_2 = 39.10 ext{ g/mol} + 2 imes 16.00 ext{ g/mol} = 71.10 ext{ g/mol}\]Similarly, for carbon dioxide ( CO _2 ):
  • Carbon ( C ) has an atomic mass of 12.01 g/mol.
Thus, the molar mass is:\[ ext{Molar mass of } CO_2 = 12.01 ext{ g/mol} + 2 imes 16.00 ext{ g/mol} = 44.01 ext{ g/mol}\]Calculating molar masses with accuracy is crucial, as these figures are used in conversions and further calculations in chemical reactions.
Stoichiometry
Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. It is based on the balanced chemical equation and allows us to predict the amounts of reactants and products involved.
To determine which reactant limits the reaction, we compare the ratio of the actual moles to the required moles from the balanced equation.
Consider the reaction involving KO _2 and CO _2 :\[ 4 KO_2(s) + 2 CO_2(g) \rightarrow 2 K_2CO_3(s) + 3 O_2(g)\]We have:
  • Moles of KO _2 = 0.0352 mol
  • Moles of CO _2 = 0.102 mol
The stoichiometric ratio requires 4 moles of KO _2 for every 2 moles of CO _2 , which can be simplified to 2:1.
The actual mole ratio calculated is 0.344.
Since this is less than the stoichiometric ratio, KO _2 is concluded as the limiting reactant.
Understanding stoichiometry is essential for recognizing which reactant will run out first and thus determines the amount of product produced.
Chemical Reaction
A chemical reaction is a process in which substances, known as reactants, are transformed into different substances, called products. These transformations are represented by chemical equations that need to be balanced to account for the conservation of mass.
In the given reaction:\[ 4 KO_2(s) + 2 CO_2(g) \rightarrow 2 K_2CO_3(s) + 3 O_2(g)\]We observe a redox reaction where the potassium superoxide ( KO _2 ) and carbon dioxide ( CO _2 ) are used to produce potassium carbonate ( K_2CO_3 ) and oxygen gas ( O_2 ).
Each atom's count matches on both sides of the equation, reflecting a balanced reaction. This balance ensures that no atoms are created or destroyed, a principle known as the Law of Conservation of Mass.
Chemical reactions are central to chemistry and reveal how substances react and transform, making them an area of great importance and interest.

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

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.