91Ó°ÊÓ

Weight before heating = 1.7839 g

Weight after heating = 1.5623 g

The amount of water lost is calculated as:

$1.7839\mathrm{g}-1.5623\mathrm{g}=0.2216\mathrm{g}=1.2301×{10}^{-2}\mathrm{mol}\mathrm{Water}$

Number of moles of water lost =$\frac{0.2216\mathrm{g}}{18\mathrm{g}}=1.2301×{10}^{-2}\mathrm{mol}$

Two moles of water is equal to one mole of barium chloride.

$\mathrm{MoleRatio}=\frac{1{\mathrm{molBaCl}}_2}{2\mathrm{mol}{\mathrm{H}}_2\mathrm{O}}=\frac{1}{2}=0.5$

Therefore, number of moles of${\mathrm{BaCl}}_2$is given below.

Molar mass of $$\begin{gathered} {\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O}=244.23\mathrm{g} \\ 0.5\left(1.2301×{10}^{-2}\right)=6.1504×{10}^{-3}\mathrm{mol}{\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

weight of 6.1504 mol$$\begin{gathered} {\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O}=6.1504×{10}^{-3}×244.23\mathrm{g} \\ =1.502\mathrm{g} \end{gathered}$$

The amount of barium and chlorine in sample is calculated as

$$\begin{gathered} \mathrm{Ba}=\left(\frac{137.327}{244.26}\right)\left(1.5023\mathrm{g}\right)=0.84462\mathrm{g} \\ \mathrm{Cl}=\left(\frac{2\left(35.452\right)}{244.26}\right)\left(1.5023\mathrm{g}\right)=0.43609\mathrm{g} \end{gathered}$$

The amount of potassium chloride present in sample is given as

1.7839g - 1.5023g = 0.2816g

$$\begin{gathered} \mathrm{K}=\left(\frac{39.098}{74.550}\right)\left(0.2816\mathrm{g}\right)=0.14769\mathrm{g} \\ \mathrm{Cl}=\left(\frac{35.452}{74550}\right)\left(0.2816\mathrm{g}\right)0.13391\mathrm{g} \end{gathered}$$

The weight percent of barium, potassium and chloride is calculated using given weight of sample.

$$\begin{gathered} \mathrm{Ba}=\frac{0.84462\mathrm{g}}{1.7839\mathrm{g}}×100\%=47.35\% \\ \mathrm{K}=\frac{0.14769\mathrm{g}}{1.7839\mathrm{g}}×100\%=8.28\% \\ \mathrm{Cl}=\frac{0.4360\mathrm{g}+0.13391\mathrm{g}}{1.7839\mathrm{g}}=31.95\% \end{gathered}$$

Conclusion

The weight percentage of barium, potassium and chloride in given sample was calculated as above.