0.1 Analyzing chemical equations  (Page 36/3)

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,

$$\text{Mass of}CaC{O}_3=0.9X500=450kg$$

The chemical reaction involved here is :

$$CaC{O}_3\to CaO+C{O}_2$$

Applying mole concept :

$$\Rightarrow \text{1 mole of}CaC{O}_3\equiv \text{1 mole of CaO}$$

$$\Rightarrow \text{(40 + 12 + 3X 16) gm of}CaC{O}_3\equiv \text{(40 + 16) gm of CaO}$$

$$\Rightarrow \text{100 gm of}CaC{O}_3\equiv \text{56 gm of CaO}$$

$$\Rightarrow \text{100 kg of}CaC{O}_3\equiv \text{56 kg of CaO}$$

Applying unitary method :

$$\Rightarrow \text{mass of CaO produced}=\frac{56X450}{100}=252kg$$

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 $\mathrm{Mn}{O}_2$ :

$$\text{Mass of}M{n}_3{O}_4\text{on conversion}=\frac{229}{261}X80=70.2gm$$

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 :

$$229\text{gm of}M{n}_3{O}_4\equiv 3X55\text{gm of Mn}\equiv 165\text{gm of Mn}$$

$$\text{Amount of Mn in 70.2 gm of}M{n}_3{O}_4=\frac{165}{229}X70.2=0.72X70.2=50.54gm$$

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

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