0.1 Analyzing chemical equations

Stoichiometry Page 1 / 3

Application of mole concept requires a balanced chemical equation. The different constituents of the reaction – reactants and products – bear a simple whole number proportion same as the proportion of the coefficients associated with constituents. According to mole concept, the molar mass of constituents participates in this proportion. For a generic consideration as given :

$x A + y B \to A_{x} B_{y}$

Here, 1 mole of compound ( $A_{x} B_{y}$ ) involves x mole of A and y mole of B. Using symbols :

$x moles of A \equiv y moles of B \equiv 1 mole of A_{x} B_{y}$

The point to emphasize here is that this is a relation, which is connected by "equivalence sign (≡)" - not by "equal to (=)" sign. We know that 2 moles of hydrogen reacts with 1 mole of oxygen to form 2 moles of water. Clearly, we can not equate like 2=1=2. We need to apply unitary method to interpret this relation of equivalence. We say that since x moles of A react with y moles of B. Hence, 1 mole of A reacts with y/x moles of "B". Similarly, 1 mole of B reacts with x/y moles of "A". Once we know the correspondence for 1 mole, we can find correspondence for any other value of participating moles of either A or B.

Mass of participating entities in a reaction

Mole concept is used to calculate mass of individual constituent of a chemical reaction. The proportion of molar mass is converted to determine proportion of mass in which entities are involved in a reaction. The symbolic mass relation for the chemical reaction as given above is :

$x M_{A} gm of A \equiv y M_{B} gm of B \equiv M_{A_{x} B_{y}} gm of A_{x} B_{y}$

We apply unitary method on the mass relation related with equivalent sign (≡) to determine mass of different entities of the reaction.

Problem : Calculate mass of lime (CaO) that can be prepared by heating 500 kg of 90 % pure limestone ( $C a C O_{3}$ .

Solution : Purity of CaCO3 is 90 %. Hence,

$Mass of C a C O_{3} = 0.9 X 500 = 450 k g$

The chemical reaction involved here is :

$C a C O_{3} \to C a O + C O_{2}$

Applying mole concept :

$\Rightarrow 1 mole of C a C O_{3} \equiv 1 mole of CaO$

$\Rightarrow (40 + 12 + 3X 16) gm of C a C O_{3} \equiv (40 + 16) gm of CaO$

$\Rightarrow 100 gm of C a C O_{3} \equiv 56 gm of CaO$

$\Rightarrow 100 kg of C a C O_{3} \equiv 56 kg of CaO$

Applying unitary method :

$\Rightarrow mass of CaO produced = \frac{56 X 450}{100} = 252 k g$

Problem : Igniting $M n O_{2}$ converts it quantitatively to $M n_{3} O_{4}$ . A sample of pyrolusite contains 80% $M n O_{2}$ , 15 % $S i O_{2}$ and 5 % water. The sample is ignited in air to constant weight. What is the percentage of manganese in the ignited sample ? ( $A_{M n} = 55$ )

Solution : The sample contains three components. Since this question involves percentage, we shall consider a sample of 100 gm. Water component weighing 5 gm evaporates on ignition. $S i O_{2}$ weighing 15 gm does not change. On the other hand, 80 gm of $M n O_{2}$ converts as :

$3 M n O_{2} \to M n_{3} O_{4} + O_{2}$

Applying mole concept,

$3 moles of M n O_{2} \equiv 1 mole of M n_{3} O_{4}$

$3 X (55 + 2 X 16) gm of M n O_{2} \equiv 1 X (3 X 55 + 4 X 16) gm of M n_{3} O_{4}$

$261 gm of M n O_{2} \equiv 229 gm of M n_{3} O_{4}$

Since sample is ignited in air to constant weight, it means that all of $Mn O_{2}$ in the sample is converted. Using unitary method, we determine mass of converted mass of $M n_{3} O_{4}$ for 80 gm of $Mn O_{2}$ :

$Mass of M n_{3} O_{4} on conversion = \frac{229}{261} X 80 = 70.2 g m$

We are required to find the percentage of Mn in the ignited sample. Thus, we need to determine the mass of the ignited sample. The ignited sample contains 70.2 gm of $M n_{3} O_{4}$ and 15 gm of $S i O_{2}$ . Total mass of ignited sample is 70.2+15 = 85.4 gm. On the other hand, amount of Mn in $M n_{3} O_{4}$ is calculated from its molecular constitution :

$229 gm of M n_{3} O_{4} \equiv 3 X 55 gm of Mn \equiv 165 gm of Mn$

$Amount of Mn in 70.2 gm of M n_{3} O_{4} = \frac{165}{229} X 70.2 = 0.72 X 70.2 = 50.54 g m$

Clearly, 85.4 gm of ignited sample contains 50.54 gm of Mn. Hence,

$Percentage amount of manganese in the ignited sample = \frac{50.54}{85.4} X 100 = 59.2 %$

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Stoichiometry. OpenStax CNX. Jul 05, 2008 Download for free at http://cnx.org/content/col10540/1.7

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Stoichiometry' conversation and receive update notifications?

Ask

	Classical Music By Marion Cabalfin Start Quiz
	1 AP 01 Human Body Anatomy Physiology Essay By OpenStax Start Flashcards
	Pre Employment English Proficiency Exam By Katherina jennife... Start Quiz
©flickr:	Biology 1 By Jill Zerressen Start Quiz
	Macroeconomics MCQ By Candice Butts Start Quiz
	Summer kitchens By Heather McAvoy Start Quiz
©flickr: Dutch	Las Vegas Timeshare By Donyea Sweets Start Test
	22 AP 22 Respiratory System Essay By OpenStax Start Flashcards
	26 AP 26 Fluid, Electrolyte, Acid-Base Balance MCQ By OpenStax Start Quiz
©flickr: Jonathan	Power By Megan Earhart Start Quiz