# 0.1 Analyzing chemical equations

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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 :

$xA+yB\to {A}_{x}{B}_{y}$

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

$\text{x moles of A}\equiv \text{y moles of B}\equiv \text{1 mole of}{A}_{x}{B}_{y}$

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 :

$x{M}_{A}\phantom{\rule{2pt}{0ex}}\text{gm of A}\equiv y{M}_{B}\phantom{\rule{2pt}{0ex}}\text{gm of B}\equiv {M}_{{A}_{x}{B}_{y}}\phantom{\rule{2pt}{0ex}}\text{gm of}{A}_{x}{B}_{y}$

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 ( $CaC{O}_{3}$ .

Solution : Purity of CaCO3 is 90 %. Hence,

$\text{Mass of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}=0.9X500=450\phantom{\rule{1em}{0ex}}kg$

The chemical reaction involved here is :

$CaC{O}_{3}\to CaO+C{O}_{2}$

Applying mole concept :

$⇒\text{1 mole of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}\equiv \text{1 mole of CaO}$

$⇒\text{(40 + 12 + 3X 16) gm of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}\equiv \text{(40 + 16) gm of CaO}$

$⇒\text{100 gm of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}\equiv \text{56 gm of CaO}$

$⇒\text{100 kg of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}\equiv \text{56 kg of CaO}$

Applying unitary method :

$⇒\text{mass of CaO produced}=\frac{56X450}{100}=252\phantom{\rule{1em}{0ex}}kg$

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

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. $Si{O}_{2}$ weighing 15 gm does not change. On the other hand, 80 gm of $Mn{O}_{2}$ converts as :

$3Mn{O}_{2}\to M{n}_{3}{O}_{4}+{O}_{2}$

Applying mole concept,

$\text{3 moles of}Mn{O}_{2}\equiv \text{1 mole of}M{n}_{3}{O}_{4}$

$3X\left(55+2X16\right)\text{gm of}Mn{O}_{2}\equiv 1X\left(3X55+4X16\right)\text{gm of}M{n}_{3}{O}_{4}$

$261\text{gm of}Mn{O}_{2}\equiv 229\text{gm of}M{n}_{3}{O}_{4}$

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

$\text{Mass of}M{n}_{3}{O}_{4}\text{on conversion}=\frac{229}{261}X80=70.2\phantom{\rule{1em}{0ex}}gm$

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 $Si{O}_{2}$ . 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 :

$229\text{gm of}M{n}_{3}{O}_{4}\equiv 3X55\text{gm of Mn}\equiv 165\text{gm of Mn}$

$\text{Amount of Mn in 70.2 gm of}M{n}_{3}{O}_{4}=\frac{165}{229}X70.2=0.72X70.2=50.54\phantom{\rule{1em}{0ex}}gm$

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

$\text{Percentage amount of manganese in the ignited sample}=\frac{50.54}{85.4}X100=59.2%$

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