Calculate its volume (in liters) of \(88.4 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) at STP.

Short Answer

Expert verified
The volume of 88.4 g of \(\mathrm{CO}_{2}\) at STP is 44.8 L.

Step by step solution

01

Determining the molar mass

First, we need to determine the molar mass of carbon dioxide (\( \mathrm{CO}_{2}\)), in terms of grams per mole (\( g/mol \)). Carbon has a molar mass of 12.01 g/mol and oxygen has a molar mass of 16.00 g/mol. Since there are two oxygen atoms for every one carbon atom in \( \mathrm{CO}_{2}\), the molar mass \( M \) is \( 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol \).
02

Calculating number of moles

Next, divide the mass given in the problem by the molar mass of \( \mathrm{CO}_{2}\) to find the number of moles \( n \). We know the mass \( m \) is 88.4 g and the molar mass \( M \) is 44.01 g/mol, so the number of moles \( n \) is \( m / M = 88.4 \, g / 44.01 \, g/mol = 2 \, moles \).
03

Calculating volume

Finally, multiply the number of moles of \( \mathrm{CO}_{2}\) by the volume that one mole of any gas occupies at STP. We know that one mole of any gas occupies a volume of 22.4 liters at STP. So, the volume \( V \) is \( n * 22.4 \, L/mol = 2 \, moles * 22.4 \, L/mol = 44.8 \, L \). So, 88.4 g of \( \mathrm{CO}_{2}\) at STP occupies a volume of 44.8 liters.

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

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