The mass of \(\mathrm{CO}_{2}\) that must be mixed with \(20 \mathrm{~g}\) of oxygen such that \(27 \mathrm{ml}\) of a sample of the resulting mixture would contain equal number of molecules of each gas (a) \(13.75 \mathrm{~g}\) (b) \(27.50 \mathrm{~g}\) (c) \(41.25 \mathrm{~g}\) (d) \(55 \mathrm{~g}\)

Short Answer

Expert verified
The mass of CO2 that must be mixed with 20 g of oxygen is 27.50 g.

Step by step solution

01

Use Avogadro's Law

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules. Since the question specifies that a sample of the resulting mixture contains an equal number of molecules of each gas, we can assume the conditions are the same and therefore, 27 ml of each gas will have the same number of molecules.
02

Calculate moles of oxygen (O2)

To find the moles of oxygen, use the molar mass of oxygen. Molecular weight of O2 is 32 g/mol. Use the formula: \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass, and \( M \) is the molar mass. For 20 g of O2, \( n = \frac{20 g}{32 g/mol} = 0.625 \) moles.
03

Equal number of moles of CO2

Since a 27 ml sample contains an equal number of molecules (and moles) of each gas, the moles of CO2 will also be 0.625.
04

Calculate mass of CO2

To find the mass of CO2, use the molar mass of CO2, which is 44 g/mol. Use the formula: \( m = n \cdot M \), where \( m \) is the mass, \( n \) is the number of moles, and \( M \) is the molar mass. For 0.625 moles, \( m = 0.625 \text{ moles} \cdot 44 \text{ g/mol} = 27.5 \text{ g} \).

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 reactants and products in a chemical reaction. It allows chemists to predict how much product can be produced from a given amount of reactant or how much reactant is required to produce a certain amount of product.

When it comes to gas mixtures, stoichiometry is closely related to Avogadro's Law, which states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. This principle is essential in solving problems where gases mix, as it provides a direct method to relate the quantities of different gases.

For instance, in the given exercise, the challenge is to balance the number of molecules of two gases, carbon dioxide (CO2) and oxygen (O2), in a mixture. The solution requires understanding the moles of gas involved, which can be calculated using stoichiometric principles along with Avogadro's Law, as demonstrated in the step-by-step solution provided.
Molar Mass Calculation
Molar mass calculation is a fundamental aspect of stoichiometry and is used to convert between the mass of a substance and the number of moles. The molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

To calculate the molar mass, one must sum the atomic masses of all the atoms in a compound, as provided by the periodic table. In our exercise, the molar mass of oxygen (\text{O}_2) is 32 g/mol, while that of carbon dioxide (CO2) is 44 g/mol. These values are essential to find the number of moles present in a given mass of substance, using the formula:
\[ n = \frac{m}{M} \]
where \( n \) represents the number of moles, \( m \) is the substance's mass, and \( M \) is the molar mass. For instance, the exercise requires calculating the mass of CO2 using its molar mass, allowing students to understand the direct proportionality between moles and mass.
Gas Mixture Composition
Gas mixture composition is essential when dealing with reactions and physical processes involving more than one gas. It requires an understanding of the individual components of the mixture and their proportions.

In the context of the exercise, understanding the gas mixture composition involves determining the mass of CO2 that must be combined with a given mass of O2 to ensure that a 27 ml sample contains equal numbers of molecules of each gas. The composition is greatly influenced by Avogadro's Law, which states that the number of molecules in a given volume of gas is constant, assuming equal temperature and pressure conditions. This forms the basis for calculating the required mass of CO2.

By utilizing stoichiometry and molar mass calculations, one can deduce the amount of each gas needed for equal molecular numbers in a sample. For the solution, equal numbers of moles of CO2 correspond to its calculated mass when mixed with 20 g of O2, thus defining the gas mixture composition.

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

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

At \(373 \mathrm{~K}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} / \mathrm{ml}\) and that of water vapour is \(0.0006 \mathrm{~g} / \mathrm{ml}\), then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) \(6 \mathrm{ml}\) (b) \(60 \mathrm{ml}\) (c) \(0.6 \mathrm{ml}\) (d) \(0.06 \mathrm{ml}\)

A mixture is made equal volume of \(\mathrm{CO}\) and air. A spark passed through so that all the oxygen is converted to carbon dioxide. What will be fractional decrease in the total volume of system assuming pressure and temperature remain constant? Air contains \(20 \%\) oxygen by volume. (a) \(0.1\) (b) \(0.2\) (c) \(0.15\) (d) \(0.3\)

Molecular mass of dry air is (a) less than moist air (b) greater than moist air (c) equal to moist air (d) may be greater or less than moist air

An amount of 1 mole of calcium cyanamide and 1 mole of water are allowed to react. The number of moles of ammonia produced is (a) \(3.0\) (b) \(2.0\) (c) \(1.0\) (d) \(0.67\)

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