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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

First we determine the limiting reagent: mass of BaCl_2 = 40.8 g ⇒ number of moles = ((40.8)/((137.3+2×35.5))) mol = 0.196 mol mass of K_2 SO_4 = 50.7 g ⇒ number of moles = ((50.7)/((2×39.0+32.0+4×16))) mol = 0.291 mol hence BaCl_2 is the limiting reagent and will determine the end of the reaction now we do a gram to gram conversion. gram → number of moles → mole ratio → gram. ((0.196 mol BaCl_2 )/1)× ((2 mol KCl)/(1 mol BaCl_2 )) × ((74.5 g KCl)/(1 mol KCl)) = 29.2 g KCl

$$\mathrm{First}\:\mathrm{we}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{reagent}: \\ $$$$\mathrm{mass}\:\mathrm{of}\:\mathrm{BaCl}_{\mathrm{2}} \:=\:\mathrm{40}.\mathrm{8}\:\mathrm{g} \\ $$$$\Rightarrow\:\mathrm{number}\:\mathrm{of}\:\mathrm{moles}\:=\:\frac{\mathrm{40}.\mathrm{8}}{\left(\mathrm{137}.\mathrm{3}+\mathrm{2}×\mathrm{35}.\mathrm{5}\right)}\:\mathrm{mol}\:=\:\mathrm{0}.\mathrm{196}\:\mathrm{mol} \\ $$$$\mathrm{mass}\:\mathrm{of}\:\mathrm{K}_{\mathrm{2}} \mathrm{SO}_{\mathrm{4}} \:=\:\mathrm{50}.\mathrm{7}\:\mathrm{g} \\ $$$$\Rightarrow\:\mathrm{number}\:\mathrm{of}\:\mathrm{moles}\:=\:\frac{\mathrm{50}.\mathrm{7}}{\left(\mathrm{2}×\mathrm{39}.\mathrm{0}+\mathrm{32}.\mathrm{0}+\mathrm{4}×\mathrm{16}\right)}\:\mathrm{mol}\:=\:\mathrm{0}.\mathrm{291}\:\mathrm{mol} \\ $$$$\mathrm{hence}\:\mathrm{BaCl}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{reagent}\:\mathrm{and}\:\mathrm{will}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{reaction} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{do}\:\mathrm{a}\:\mathrm{gram}\:\mathrm{to}\:\mathrm{gram}\:\mathrm{conversion}. \\ $$$$\mathrm{gram}\:\rightarrow\:\mathrm{number}\:\mathrm{of}\:\mathrm{moles}\:\rightarrow\:\mathrm{mole}\:\mathrm{ratio}\:\rightarrow\:\mathrm{gram}. \\ $$$$\:\frac{\mathrm{0}.\mathrm{196}\:\cancel{\mathrm{mol}\:\mathrm{BaCl}_{\mathrm{2}} }}{\mathrm{1}}×\:\frac{\mathrm{2}\:\cancel{\mathrm{mol}\:\mathrm{KCl}}}{\mathrm{1}\:\cancel{\mathrm{mol}\:\mathrm{BaCl}_{\mathrm{2}} }}\:×\:\frac{\mathrm{74}.\mathrm{5}\:\mathrm{g}\:\mathrm{KCl}}{\mathrm{1}\:\cancel{\mathrm{mol}\:\mathrm{KCl}}}\:=\:\mathrm{29}.\mathrm{2}\:\mathrm{g}\:\mathrm{KCl} \\ $$

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

I appreciate sir. Thanks

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

Commented by Tawa11 last updated on 14/Sep/21

nice

$$\mathrm{nice} \\ $$

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