0.1 Analyzing chemical equations  (Page 3/3)

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

Solution : The chemical reaction involved in the conversion is :

$$Fe+S\to FeS$$

$$\text{1 mole of Fe}\equiv \text{1 moles of S}$$

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 :

$$n_{Fe}=\frac{g}{56}$$

and

$$y{n}_{Fe}=\frac{g}{56}$$

Similarly,

$$x{n}_S=\frac{g}{32}$$

Here, $y{n}_{Fe}<x{n}_S$ . 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.

$$\Rightarrow \text{Mass of Sulphur converted in the reaction}=\frac{x}{56}X{M}_S=\frac{32x}{56}$$

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

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 ( $AgN{O}_3$ ). 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 :

$$KCl+AgN{O}_3\to KN{O}_3+AgCl$$

$$KI+AgN{O}_3\to KN{O}_3+AgI$$

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

$$\text{1 mole of KCl}\equiv \text{1 mole of AgCl}$$

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

Therefore, mass of AgCl is :

$$\frac{x}{74.5}\text{moles of AgCl}=\frac{x}{74.5}X\left(108+35.5\right)gm=\frac{143.5x}{74.5}gm$$

Similarly, applying mole concept to second equation,

$$\text{1 mole of KI}\equiv \text{1 mole of AgI}$$

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

Therefore, mass of AgI is :

$$\frac{y}{166}\text{moles of AgI}=\frac{y}{166}X\left(108+127\right)gm=235\frac{y}{166}gm$$

According to question,

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

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

$$x+\frac{74.5X235y}{143.5X166}=\frac{3.8X74.5}{143.5}$$

$$\Rightarrow x+0.73y=1.97$$

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

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

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

$$\Rightarrow 0.27x=0.36$$

$$\Rightarrow x=1.34gm$$

$$\Rightarrow y=2.2-x=2.2-1.34=0.86gm$$

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

Solution : Here,

$$\Rightarrow \text{1 mole of}Fe{S}_2\equiv \text{2 moles of}{H}_{2}S{O}_4$$

$$\Rightarrow \text{(56+2X32) gm of}Fe{S}_2\equiv \text{2X(2X1+32+4X16) moles of}{H}_{2}S{O}_4$$

$$\Rightarrow \text{120 gm of}Fe{S}_2\equiv \text{196 gm of}{H}_{2}S{O}_4$$

Applying unitary method,

$$\Rightarrow \text{mass of}{H}_{2}S{O}_4=\frac{192}{120}X100=160gm$$