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Question Number 132337 by I want to learn more last updated on 13/Feb/21

Answered by physicstutes last updated on 14/Feb/21

$$\mathrm{First}\mathrm{we}\mathrm{determine}\mathrm{the}\mathrm{limiting}\mathrm{reagent}:$$$$\mathrm{mass}\mathrm{of}{\mathrm{BaCl}}_2=40.8\mathrm{g}$$$$\Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{40.8}{(137.3+2\times 35.5)}\mathrm{mol}=0.196\mathrm{mol}$$$$\mathrm{mass}\mathrm{of}{\mathrm{K}}_2{\mathrm{SO}}_4=50.7\mathrm{g}$$$$\Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{50.7}{(2\times 39.0+32.0+4\times 16)}\mathrm{mol}=0.291\mathrm{mol}$$$$\mathrm{hence}{\mathrm{BaCl}}_2\mathrm{is}\mathrm{the}\mathrm{limiting}\mathrm{reagent}\mathrm{and}\mathrm{will}\mathrm{determine}\mathrm{the}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{reaction}$$$$\mathrm{now}\mathrm{we}\mathrm{do}\mathrm{a}\mathrm{gram}\mathrm{to}\mathrm{gram}\mathrm{conversion}.$$$$\mathrm{gram}\to \mathrm{number}\mathrm{of}\mathrm{moles}\to \mathrm{mole}\mathrm{ratio}\to \mathrm{gram}.$$$$\frac{0.196\cancel{\mathrm{mol}{\mathrm{BaCl}}_2}}{1}\times \frac{2\cancel{\mathrm{mol}\mathrm{KCl}}}{1\cancel{\mathrm{mol}{\mathrm{BaCl}}_2}}\times \frac{74.5\mathrm{g}\mathrm{KCl}}{1\cancel{\mathrm{mol}\mathrm{KCl}}}=29.2\mathrm{g}\mathrm{KCl}$$

Commented by I want to learn more last updated on 23/Feb/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{Thanks}$$

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{nice}$$

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