InterviewSolution
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InterviewSolution
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Upon mixming `100.0 mL` to `0.1 M` potassium solphate solution and `100.0 mL` of `0.05 M` barium chloride solution, precipitation of barium sulphate takes place. How many moles of barium sulphate are formed? Also, calculate the molar concentration of species left behind in the solution. Which is the limiting reagent? |
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Answer» `K_(2) SO_(4) + BaCl_(2) rarr BaSO_(4) darr + 2 KCl` `(Acc.to equation), 1 "mmol, 1mmol, 1mmol, 2mmol")`, equation), mmole before, `10 xx 01, 100 xx 0.05`, -, - (reaction, = 10 mmol, = mmol), mmole left after, 10 - 5, -, 5 mmol, 10 mmol), reaction), = 5 mmol 5 mmol of `BaCl_(2)` will react will 5 mmol of `K_(2) SO_(4)`. Hence`BaCl_(2)` is the limiting reagent which is completely consumed in the reaction and it will decied the number of millimiles of products formed. Total volume of solution `= 100 + 100 = 200 mL` Moles of `BaSO_(4)` precipitated `= (5 mmol)/(1000 mL) = 0.005 mol` Unreacted `K_(2) SO_(4)` and `KCl` are left in the solution. Concentration of `K_(2) SO_(4)` and `KCl` are left in the solution `= (5 mmol)/(200 mL)` `= 0.025 M` Concentration of `KCl` formed in the reaction `= (10 mmol)/(200 mL)` `= 0.05 M` |
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