Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 126

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess HCl gas to produce \(3.22 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at \(373 \mathrm{~K}\) and \(0.951\) atm and a manganese chloride compound \(\left(\mathrm{MnCl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?

Short Answer

Expert verified
The formula of the manganese chloride compound produced in the reaction is MnClx, where \(x\) is determined by dividing the moles of chlorine by the moles of manganese (\(x\) = moles of Cl / moles of Mn). Round the value of \(x\) to the nearest whole number to find the whole number ratio of chlorine atoms interacting with manganese atoms in the compound.

Step by step solution

01

Calculate the moles of hydrogen gas produced

We can use the Ideal Gas Law to determine the moles of hydrogen gas (\(H_2\)) produced: \[PV = nRT\] Where \(P\) is the pressure in atm = \(0.951 \mathrm{~atm}\), \(V\) is the volume in L = \(3.22 \mathrm{~L}\), \(n\) is the moles of hydrogen gas, which we need to find, \(R\) is the gas constant = \(0.0821 \frac{\mathrm{L \cdot atm} }{ \mathrm{mol \cdot K}}\), \(T\) is the temperature in Kelvin = \(373 \mathrm{~K}\). We rearrange the equation to solve for \(n\): \[n = \frac{PV}{RT}\] Now plug in the values: \[n = \frac{(0.951 \mathrm{~atm})(3.22 \mathrm{~L})}{(0.0821 \frac{\mathrm{L \cdot atm} }{ \mathrm{mol \cdot K}})(373 \mathrm{~K})}\]
02

Calculate the moles of manganese in the sample

We know the mass of manganese in the sample and its molar mass (Mn: \(54.94 \frac{\mathrm{g}}{\mathrm{mol}}\)), so we can find the moles of manganese: Moles of Mn = \(\frac{\text{Mass of Mn}}{\text{Molar Mass of Mn}}\) Moles of Mn = \(\frac{2.747 \mathrm{~g}}{54.94 \frac{\mathrm{g}}{\mathrm{mol}}}\)
03

Calculate the moles of chlorine in the manganese chloride compound

Using stoichiometry, we know that the reaction is as follows: Mn(s) + xHCl(g) -> MnClx(s) + x/2 H2(g) From the moles of hydrogen calculated in the previous step, we can find out the moles of chlorine present in the sample: Moles of Cl = 2 * moles of H2 (from the stoichiometry of the balanced equation)
04

Deduce the formula of manganese chloride

To determine the formula of manganese chloride, divide the moles of chlorine by moles of manganese: \(x\) = moles of Cl / moles of Mn We need to round the value of \(x\) to the nearest whole number because we are looking for a whole number ratio of chlorine atoms interacting with manganese atoms. This whole number will give us the formula of manganese chloride as MnClx.

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

The steel reaction vessel of a bomb calorimeter, which has a volume of \(75.0 \mathrm{~mL}\), is charged with oxygen gas to a pressure of \(14.5\) atm at \(22^{\circ} \mathrm{C}\). Calculate the moles of oxygen in the reaction vessel.

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advan- tages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{~g}\) argon gas. The pressure inside the cylinder is \(2050 . \mathrm{psi}\) (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to 650 . psi at a temperature of \(26^{\circ} \mathrm{C}\) ?

A 2.50-L container is filled with \(175 \mathrm{~g}\) argon. a. If the pressure is \(10.0 \mathrm{~atm}\), what is the temperature? b. If the temperature is \(225 \mathrm{~K}\), what is the pressure?

Solid thorium(IV) fluoride has a boiling point of \(1680^{\circ} \mathrm{C}\). What is the density of a sample of gaseous thorium(IV) fluoride at its boiling point under a pressure of \(2.5\) atm in a 1.7-L container? Which gas will effuse faster at \(1680^{\circ} \mathrm{C}\), thorium(IV) fluoride or uranium(III) fluoride? How much faster?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

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