Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 110

A \(1.50 \mathrm{~g}\) sample of potassium bicarbonate having \(80 \%\) purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is (a) \(3.72 \mathrm{~g}\) (b) \(0.72 \mathrm{~g}\) (c) \(0.372 \mathrm{~g}\) (d) \(0.186 \mathrm{~g}\)

Short Answer

Expert verified
a) 0.299 g b) 0.345 g c) 0.372 g d) 0.400 g Answer: c) 0.372 g

Step by step solution

01

Find the amount of potassium bicarbonate in the sample

Since we have \(80 \%\) purity, the amount of \(\mathrm{KHCO_3}\) in the sample can be found using the following formula: $$\mathrm{Amount~(pure) = total~amount \cdot percentage~purity}$$ $$\mathrm{Amount~(pure) = 1.50~g \cdot 0.80}$$ $$\mathrm{Amount~(pure) = 1.20~g}$$ So, there is \(1.20 \mathrm{~g}\) of \(\mathrm{KHCO_3}\) in the sample.
02

Find the moles of potassium bicarbonate

Now, we will find the moles of \(\mathrm{KHCO_3}\) using the molar mass: $$\mathrm{Moles~of~KHCO_3 = \frac{mass}{molar~mass}}$$ The molar mass of \(\mathrm{KHCO_3}\) can be found as follows: \(\mathrm{K: ~39.10~g/mol}\), \(\mathrm{H: ~1.01~g/mol}\), \(\mathrm{C: ~12.01~g/mol}\), and \(\mathrm{O: ~16.00~g/mol}\) \(\mathrm{Molar ~mass ~of ~KHCO_3 = (39.10 + 1.01 + 12.01 + 3 \cdot 16.00) ~g/mol = 100.12 ~g/mol}\) So, $$\mathrm{Moles~of~KHCO_3 = \frac{1.20 ~g}{100.12 ~g/mol}}$$ \(\mathrm{Moles~of~KHCO_3 = 0.01198 ~mol}\) (approx.)
03

Calculate the moles of water and carbon dioxide formed

Referring back to the balanced equation, $$\mathrm{2KHCO_3 \rightarrow K_2CO_3 + H_2O + CO_2}$$ From the stoichiometry of the reaction, we see that \(1\) mole of \(\mathrm{H_2O}\) and \(1\) mole of \(\mathrm{CO_2}\) are produced for every \(2\) moles of \(\mathrm{KHCO_3}\). Using this information, we can find the moles of \(\mathrm{H_2O}\) and \(\mathrm{CO_2}\) formed as follows: $$\mathrm{Moles~of~H_2O = \frac{1}{2} \cdot \mathrm{moles~of~KHCO_3}}$$ $$\mathrm{Moles~of~H_2O = \frac{1}{2} \cdot 0.01198 ~mol \approx 0.00599 ~mol}$$ $$\mathrm{Moles~of~CO_2 = \frac{1}{2} \cdot \mathrm{moles~of~KHCO_3}}$$ $$\mathrm{Moles~of~CO_2 = \frac{1}{2} \cdot 0.01198 ~mol \approx 0.00599 ~mol}$$
04

Calculate the mass of water and carbon dioxide formed

Now, we'll find the mass of \(\mathrm{H_2O}\) and \(\mathrm{CO_2}\) formed using their molar masses: Molar mass of \(\mathrm{H_2O}: 18.02 ~g/mol\) Molar mass of \(\mathrm{CO_2}: 44.01 ~g/mol\) $$\mathrm{Mass~of~H_2O = moles~of~H_2O \cdot molar~mass~of~H_2O}$$ $$\mathrm{Mass~of~H_2O = 0.00599 ~mol \cdot 18.02 ~g/mol \approx 0.1079 ~g}$$ $$\mathrm{Mass~of~CO_2 = moles~of~CO_2 \cdot molar~mass~of~CO_2}$$ $$\mathrm{Mass~of~CO_2 = 0.00599 ~mol \cdot 44.01 ~g/mol \approx 0.2637 ~g}$$
05

Calculate the loss in weight of the sample

The total mass of \(\mathrm{H_2O}\) and \(\mathrm{CO_2}\) lost during the heating is the sum of their masses: $$\mathrm{Loss~in~weight = mass~of~H_2O + mass~of~CO_2}$$ $$\mathrm{Loss~in~weight \approx 0.1079 ~g + 0.2637 ~g}$$ $$\mathrm{Loss~in~weight \approx 0.372 ~g}$$ So, the loss in weight of the sample on heating is approximately \(0.372 \mathrm{~g}\), which corresponds to the option (c).

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

For \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\), which is the correct mole relationship? (a) \(9 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}\) (b) \(5 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}\) (c) \(9 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}_{2}\) (d) mole of \(\mathrm{Cu}=5 \times\) mole of \(\mathrm{O}\)

An organic compound contains \(40 \%\) carbon and \(6.67 \%\) hydrogen by mass. Which of the following represents the empirical formula of the compound? (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{2} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{O}\)

From \(2 \mathrm{mg}\) calcium, \(1.2 \times 10^{19}\) atoms are removed. The number of g-atoms of calcium left is \((\mathrm{Ca}=40)\) (a) \(5 \times 10^{-3}\) (b) \(2 \times 10^{-3}\) (c) \(3 \times 10^{-3}\) (d) \(5 \times 10^{-6}\)

At room temperature, the molar volume of hydrogen fluoride gas has a mass of about \(50 \mathrm{~g}\). The formula weight of hydrogen fluoride is 20 . Gaseous hydrogen fluoride at room temperature is therefore, probably a mixture of (a) \(\mathrm{H}_{2}\) and \(\mathrm{F}_{2}\) (b) \(\mathrm{HF}\) and \(\mathrm{H}_{2} \mathrm{~F}_{2}\) (c) \(\mathrm{HF}\) and \(\mathrm{H}_{2, s} \mathrm{~F}_{2.5}\) (d) \(\mathrm{H}_{2} \mathrm{~F}_{2}\) and \(\mathrm{H}_{3} \mathrm{~F}_{3}\)

If rocket were fuelled with kerosene and liquid oxygen, what mass of oxygen would be required for every litre of kerosene? Assume kerosene to have the average composition \(\mathrm{C}_{14} \mathrm{H}_{30}\) and density, \(0.792 \mathrm{~g} / \mathrm{ml}\). (a) \(5.504 \mathrm{~kg}\) (b) \(2.752 \mathrm{~kg}\) (c) \(1.376 \mathrm{~kg}\) (d) \(3.475 \mathrm{~kg}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Inorganic 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