Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 91

When \(20 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) is reacted with \(50 \mathrm{~g}\) of \(\mathrm{HCl}, \mathrm{FeCl}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed. The amount of unreacted \(\mathrm{HCl}\) is \((\mathrm{Fe}=56)\) (a) \(27.375 \mathrm{~g}\) (b) \(22.625 \mathrm{~g}\) (c) \(30 \mathrm{~g}\) (d) \(4.75 \mathrm{~g}\)

Short Answer

Expert verified
Answer: Approximately 22.625 g of unreacted HCl is left after the reaction.

Step by step solution

01

Write the balanced chemical equation

The chemical reaction between Fe₂O₃ and HCl is as follows: Fe₂O₃ (s) + 6HCl (aq) -> 2FeCl₃ (aq) + 3H₂O (l)
02

Calculate the moles of Fe₂O₃ and HCl

First, we need to calculate the molar mass of each compound involved: Fe₂O₃: (2 × 56) + (3 × 16) = 112 + 48 = 160 g/mol HCl: (1 × 1) + (1 × 35.5) = 1 + 35.5 = 36.5 g/mol Now, we can calculate the moles of each reactant: moles of Fe₂O₃ = mass of Fe₂O₃ / molar mass of Fe₂O₃ = 20 g / 160 g/mol = 0.125 mol moles of HCl = mass of HCl / molar mass of HCl = 50 g / 36.5 g/mol = 1.369 mol
03

Determine the limiting reactant and the excess reactant

To find out the limiting reactant, we need to compare the mole ratios of the reactants from the balanced chemical equation. Mole ratio of Fe₂O₃ to HCl: 1:6 Reaction ratio of Fe₂O₃ to HCl: 0.125mol : 1.369mol Now, we need to determine the number of moles of HCl that would be required for the given amount of Fe₂O₃: required moles of HCl = moles of Fe₂O₃ × 6 = 0.125 mol × 6 = 0.75 mol Since we have 1.369 mol of HCl, which is more than the required 0.75 mol for the reaction, the limiting reactant is Fe₂O₃, and the excess reactant is HCl.
04

Calculate the amount of unreacted HCl

Now that we know that Fe₂O₃ is the limiting reactant, we can determine the amount of unreacted HCl: unreacted moles of HCl = moles of HCl - moles of required HCl = 1.369 mol - 0.75 mol = 0.619 mol Now we convert moles of unreacted HCl to grams: mass of unreacted HCl = unreacted moles of HCl × molar mass of HCl = 0.619 mol × 36.5 g/mol = 22.594 g So the amount of unreacted HCl is approximately 22.625 g, which is the answer (b).

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 density (in \(\mathrm{g} /\) l) of an equimolar mixture of methane and ethane at 1 atm and \(0^{\circ} \mathrm{C}\) is (a) \(1.03\) (b) \(2.05\) (c) \(0.94\) (d) \(1.25\)

An ore contains \(2.296 \%\) of the mineral argentite, \(\mathrm{Ag}_{2} \mathrm{~S}\), by mass. How many grams of this ore would have to be

Samples of \(1.0 \mathrm{~g}\) of \(\mathrm{Al}\) are treated separately with an excess of sulphuric acid and an excess of sodium hydroxide. The ratio of the number of moles of the hydrogen gas evolved is (a) \(1: 1\) (b) \(3: 2\) (c) \(2: 1\) (d) \(9: 4\)

A compound contains elements \(X\) and \(\mathrm{Y}\) in \(1: 4\) mass ratio. If the atomic masses of \(X\) and \(Y\) are in \(1: 2\) ratio, the empirical formula of compound should be (a) \(\mathrm{XY}_{2}\) (b) \(\mathrm{X}_{2} \mathrm{Y}\) (c) \(\mathrm{XY}_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)

Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is \(69.98 \%\). What is the molecular mass of cortisone? (a) \(180.05\) (b) \(360.1\) (c) \(312.8\) (d) \(205.8\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical 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