0.1 Analyzing chemical equations (Page 3/3)

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Limiting reactant

The participating mass of the reactants may not be exactly same as required by balanced chemical reaction. The reactant which limits the progress of reaction is called limiting reactant (reagent). If gA and gB be the mass of two reactants, then moles present are :

$n_{A} = \frac{g_{A}}{M_{A}}$

$n_{B} = \frac{g_{B}}{M_{B}}$

According to mole concept :

$x moles of A \equiv y moles of B$

$nA moles of A \equiv \frac{y n_{A}}{x} moles of B$

For reaction to complete as the limiting case,

$\frac{y n_{A}}{x} = n_{B}$

$y n_{A} = x n_{B}$

For A to be limiting,

$n_{B} > \frac{y n_{A}}{x}$

$y n_{A} < x n_{B}$

For B to be limiting,

$n_{B} < \frac{y n_{A}}{x}$

$x n_{B} < y n_{A}$

$y n_{A} > x n_{B}$

Problem : Equal amounts of iron and sulphur are heated together to form FeS. What fraction of iron or Sulphur are converted into FeS. Here, $A_{F e} = 56$ and $A_{S} = 32$ .

Solution : The chemical reaction involved in the conversion is :

$F e + S \to F e S$

$1 mole of Fe \equiv 1 moles of S$

Here, x=y=1. Now, equal amounts are used. Let “g” gm of each takes part in the reaction. The numbers of moles corresponding to “g” gm are :

$n_{F e} = \frac{g}{56}$

and

$y n_{F e} = \frac{g}{56}$

Similarly,

$x n_{S} = \frac{g}{32}$

Here, $y n_{F e} < x n_{S}$ . It means that iron is in short supply and is the limiting reactant. Clearly, all iron is converted into FeS. Hence, fraction of iron used is 1. On the other hand, the moles of Sulphur converted are $\frac{x}{56}$ only as reaction is limited by iron.

$\Rightarrow Mass of Sulphur converted in the reaction = \frac{x}{56} X M_{S} = \frac{32 x}{56}$

$\Rightarrow Fraction of suphur converted = 32 x / 56 x = \frac{32}{56} = 0.571$

Multiple reactions

Some chemical reaction under analysis involves simultaneous or concurrent reactions. Consider the example in which a mixture of KCl and KI are treated with excess of silver nitrate ( $A g N O_{3}$ ) :

$K C l + A g N O_{3} \to K N O_{3} + A g C l$

$K I + A g N O_{3} \to K N O_{3} + A g I$

We apply mole concept to two concurrent reactions as :

$1 mole of KCl \equiv 1 mole of A g N O_{3} \equiv 1 mole of K N O_{3} \equiv 1 mole of AgCl$

and

$1 mole of KI \equiv 1 mole of A g N O_{3} \equiv 1 mole of K N O_{3} \equiv 1 mole of AgI$

Problem : 2.2 g of a mixture of KCl and KI yields 3.8 gm of AgCl and AgI, when treated with excess silver nitrate ( $A g N O_{3}$ ). Find the mass of KCl and KI in the original mixture.

Solution : Since the mixture is treated with excess silver nitrate, both chloride and iodide are completely consumed, forming silver chloride and iodide. Let the initial mixture contains x and y grams of KCl and KI respectively. Then, according to question :

$x + y = 2.2$

This is first of two linear equations, which will be used to determine x and y. Now, chemical reactions are :

$K C l + A g N O_{3} \to K N O_{3} + A g C l$

$K I + A g N O_{3} \to K N O_{3} + A g I$

The amount of AgCl and AgI produced is 3.8 gm. Applying mole concept to first equation,

$1 mole of KCl \equiv 1 mole of AgCl$

$\frac{x}{(39 + 35.5)} moles of KCl \equiv \frac{x}{(39 + 35.5)} moles of AgCl \equiv \frac{x}{74.5} moles of AgCl$

Therefore, mass of AgCl is :

$\frac{x}{74.5} moles of AgCl = \frac{x}{74.5} X (108 + 35.5) g m = \frac{143.5 x}{74.5} g m$

Similarly, applying mole concept to second equation,

$1 mole of KI \equiv 1 mole of AgI$

$\frac{x}{(39 + 127)} moles of KCl \equiv \frac{y}{(39 + 127)} moles of AgI \equiv \frac{y}{166} moles of AgI$

Therefore, mass of AgI is :

$\frac{y}{166} moles of AgI = \frac{y}{166} X (108 + 127) g m = 235 \frac{y}{166} g m$

According to question,

$\frac{143.5 x}{74.5} + \frac{235 y}{166} = 3.8$

In order to render coefficient of first term of the equation equal to 1 and hence simplify the equation, we multiply the equation by $\frac{74.5}{143.5}$ ,

$x + \frac{74.5 X 235 y}{143.5 X 166} = \frac{3.8 X 74.5}{143.5}$

$\Rightarrow x + 0.73 y = 1.97$

Thus, we have two linear equations and two unknowns. Substituting for y in terms of x from first linear equation derived earlier, we have :

$\Rightarrow x + 0.73 (2.2 - x) = 1.97$

$\Rightarrow x + 1.61 - 0.73 x = 1.97$

$\Rightarrow 0.27 x = 0.36$

$\Rightarrow x = 1.34 g m$

$\Rightarrow y = 2.2 - x = 2.2 - 1.34 = 0.86 g m$

Chain reaction

A chemical process involves multiple reactions in which product of one reaction becomes reactant in the second reaction. Consider example of formation of sulphuric acid from pyrite( $F e S_{2}$ ) :

$4 F e S_{2} + 11 O_{2} \to 2 F e_{2} O_{3} + 8 S O_{2}$

$2 S O_{2} + O_{2} \to 2 S O_{3}$

$S O_{3} + H_{2} O \to H_{2} S O_{4}$

Here, first reaction supplies $S O_{2}$ required as reactant in the second reaction. We, therefore, need to multiply second reaction 4 times so that $S O_{2}$ molecules are balanced. Similarly, third reaction should be multiplied by 8 to match $8 S O_{3}$ molecules produced in the second reaction. Thus, balanced chain reaction is :

$4 F e S_{2} + 11 O_{2} \to 2 F e_{2} O_{3} + 8 S O_{2}$

$8 S O_{2} + 4 O_{2} \to 8 S O_{3}$

$8 S O_{3} + 8 H_{2} O \to 8 H_{2} S O_{4}$

From balanced chain equations, we conclude that :

$\Rightarrow 4 moles of F e S_{2} \equiv 8 moles of H_{2} S O_{4}$

$\Rightarrow 1 mole of F e S_{2} \equiv 2 moles of H_{2} S O_{4}$

Alternatively, we can argue that one molecule of $F e S_{2}$ contains 2 atoms of Sulphur(S). On the other hand, one molecule of $H_{2} S O_{4}$ contains 2 atoms of Sulphur(S). Further, there is no loss or gain of Sulphur in chain reactions. Hence, to conserve mass of suphur,

$\Rightarrow 1 mole of F e S_{2} \equiv 2 moles of H_{2} S O_{4}$

Problem : How many grams of sulphuric acid can be obtained from 100 gm of pyrite?

Solution : Here,

$\Rightarrow 1 mole of F e S_{2} \equiv 2 moles of H_{2} S O_{4}$

$\Rightarrow (56+2X32) gm of F e S_{2} \equiv 2X(2X1+32+4X16) moles of H_{2} S O_{4}$

$\Rightarrow 120 gm of F e S_{2} \equiv 196 gm of H_{2} S O_{4}$

Applying unitary method,

$\Rightarrow mass of H_{2} S O_{4} = \frac{192}{120} X 100 = 160 g m$

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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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