Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 8

Fill in the blanks. The ratio of the volumes of \(11 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28 \mathrm{~g}\) of \(\mathrm{CO}\) at \(\mathrm{STP}\) is

Short Answer

Expert verified
Answer: The ratio of the volumes of 11 g of CO₂ and 28 g of CO at STP is 7:1.

Step by step solution

01

Calculate the number of moles of each gas

First, we need to find the number of moles of each gas. We can do this by dividing their given masses by their molar masses. The molar mass of \(\mathrm{CO}_{2}\) is 44 g/mol and the molar mass of \(\mathrm{CO}\) is 28 g/mol. Number of moles of \(\mathrm{CO}_{2}\) = \(\frac{11\mathrm{~g}}{44\mathrm{~g/mol}}\) Number of moles of \(\mathrm{CO}\) = \(\frac{28\mathrm{~g}}{28\mathrm{~g/mol}}\)
02

Calculate the volumes of each gas

Now, we will calculate the volumes of each gas using the number of moles and the molar volume of an ideal gas at STP (22.4 liters/mol). Volume of \(\mathrm{CO}_{2}\) = Number of moles of \(\mathrm{CO}_{2}\) × Molar volume at STP Volume of \(\mathrm{CO}_{2}\) = \(\frac{11\mathrm{~g}}{44\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}\) Volume of \(\mathrm{CO}\) = Number of moles of \(\mathrm{CO}\) × Molar volume at STP Volume of \(\mathrm{CO}\) = \(\frac{28\mathrm{~g}}{28\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}\)
03

Find the ratio of the volumes

Finally, we will find the ratio of the volumes of the two gases. We can do this by dividing the volume of \(\mathrm{CO}_{2}\) by the volume of \(\mathrm{CO}\). Ratio = \(\frac{\text{Volume of }\mathrm{CO}_{2}}{\text{Volume of }\mathrm{CO}}\) Ratio = \(\frac{\frac{11\mathrm{~g}}{44\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}}{\frac{28\mathrm{~g}}{28\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}}\) Ratio = \(\frac{11}{44} \times \frac{28}{1}=\boxed{7}\) So, the ratio of the volumes of \(11 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28 \mathrm{~g}\) of \(\mathrm{CO}\) at STP is 7:1.

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

Fill in the blanks. The volume of \(\mathrm{CO}_{2}\) liberated at \(\mathrm{STP}\) is on thermal decomposition of \(84 \mathrm{~g}\) of sodium bicarbonate.

Select the correct alternative from the given choices. \(20 \mathrm{cc}\) of a hydrocarbon on complete combustion gave \(80 \mathrm{cc}\) of \(\mathrm{CO}_{2}\) and \(100 \mathrm{cc}\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(\mathrm{STP}\). The empirical formula of that compound is (a) \(\mathrm{C}_{2} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10}\)

If \(15 \mathrm{mg}\) of \(\mathrm{N}_{2} \mathrm{O}_{3}\) is added to \(4.82 \times 10^{20}\) molecules of \(\mathrm{N}_{2} \mathrm{O}_{3}\) the total volume occupied by the gas at STP is (a) \(0.044 \ell\) (b) \(0.022 \ell\) (c) \(0.22 \ell\) (d) \(0.44 \ell\)

State whether the following statements are true or false. The greater the critical temperature of a gas, the easier is the liquefaction of the gas.

Select the correct alternative from the given choices. When \(180 \mathrm{~g}\) of glucose is subjected to combustion, the volume of \(\mathrm{CO}_{2}\) liberated at \(\mathrm{STP}\) is (a) \(22.4 \ell\) (b) \(67.2\) \ell (c) \(44 \ell\) (d) \(134.4 \ell\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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