Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 65

Write a balanced equation for the reaction of hydrofluoric acid with \(\mathrm{SiO}_{2}\). What volume of \(2.0 \mathrm{M} \mathrm{HF}\) is required to react with one gram of silicon dioxide?

Short Answer

Expert verified
Short Answer: The balanced equation for the reaction between hydrofluoric acid (HF) and silicon dioxide (SiO2) is SiO2 + 6 HF → H2SiF6 + 2 H2O. To react with 1 gram of silicon dioxide, approximately 49.8 mL of a 2.0 M HF solution is required.

Step by step solution

01

Write a balanced equation for the reaction

The reaction between hydrofluoric acid (HF) and silicon dioxide (SiO2) produces hexafluorosilicic acid (H2SiF6) and water (H2O). The balanced equation for the reaction is: \(\mathrm{SiO_{2}} + 6 \mathrm{HF} \rightarrow \mathrm{H_{2}SiF_{6}}+ 2 \mathrm{H_{2}O}\)
02

Calculate the amount of moles of \(\mathrm{SiO}_{2}\) in 1 gram

In order to determine the amount of moles of \(\mathrm{SiO}_{2}\), we have to divide the given mass of \(\mathrm{SiO}_{2}\) by its molar mass. The molar mass of \(\mathrm{SiO}_{2}\) is: \(\mathrm{SiO_{2}} = (1 \times 28.09 \thinspace g/mol (Si) + 2 \times 16.00 \thinspace g/mol (O)) = 60.09 \thinspace g/mol\) So, the amount of moles of \(\mathrm{SiO}_{2}\) in 1 gram is: \(\mathrm{1 \thinspace gram} \times \frac{1}{60.09 \thinspace g/mol} = 0.0166 \thinspace mol\)
03

Calculate the amount of moles of HF required to react with 1 gram of \(\mathrm{SiO}_{2}\)

From the balanced equation, the mole ratio between \(\mathrm{SiO}_{2}\) and \(\mathrm{HF}\) is 1:6. Therefore, to find the amount of moles of \(\mathrm{HF}\) required, we can multiply the amount of moles of \(\mathrm{SiO}_{2}\) by the mole ratio: \(0.0166 \thinspace mol \textrm{ of SiO}_{2} \times \frac{6 \thinspace mol \textrm{ of HF}}{1 \thinspace mol \textrm{ of SiO}_{2}} = 0.0996 \thinspace mol\)
04

Calculate the volume of a 2.0 M HF solution required

The concentration of the HF solution is given as 2.0 M (moles per liter). To find the volume of the solution required to provide 0.0996 moles of HF, we can use the following equation: \(\mathrm{Volume} = \frac{\mathrm{moles}}{\mathrm{concentration}} = \frac{0.0996 \thinspace mol}{2.0 \thinspace M} = 0.0498 \thinspace L\) Finally, convert the volume to milliliters: \(\mathrm{0.0498 \thinspace L} \times \frac{1000 \thinspace mL}{1 \thinspace L} = 49.8 \thinspace mL\) So, the volume of 2.0 M HF required to react with 1 gram of silicon dioxide is approximately 49.8 mL.

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

Give the formula for the acidic oxide of (a) \(\mathrm{HNO}_{3}\) (b) \(\mathrm{HNO}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Give the Lewis structure of (a) \(\mathrm{NO}_{2}\) (b) \(\mathrm{NO}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{SO}_{3}\)

Write a balanced equation for the reaction of ammonia with (a) \(\mathrm{Cu}^{2+}\) (b) \(\mathrm{H}^{+}\) (c) \(\mathrm{Al}^{3+}\)

Write a balanced net ionic equation for the disproportionation reaction (a) of iodine to give iodate and iodide ions in basic solution. (b) of chlorine gas to chloride and perchlorate ions in basic solution.

If an electrolytic cell producing fluorine uses a current of \(7.00 \times 10^{3} \mathrm{~A}\) (at \(10.0 \mathrm{~V})\), how many grams of fluorine gas can be produced in two days (assuming that the cell operates continuously at \(95 \%\) efficiency)?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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