A 0.5087 g solid sample containing a mixture of LaCl3 (molar mass = 245.26 g/mol

Question

A 0.5087 g solid sample containing a mixture of LaCl3 (molar mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was dissolved in water. The solution was titrated with KIO3, producing the precipitates La(IO3)3(s) and Ce(IO3)3(s). For the complete titration of both La3 and Ce3 , 44.33 mL of 0.1274 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.

Question 12 of 21 Incorrect X Incorrect Incorrect X Incorrect Incorrect X Map Sapling Learning macmillan learning A 0.5087 g solid sample containing a mixture of LaCl3 (molar mass s 245.26 g/mol) and Ce(NO3)3 (molar mass 326.13 g/mol) was dissolved in water. The solution was titrated with KIO3, producing the precipitates La(IO3)3(s) and Ce(IO3)3(s). For the complete titration of both 3+ and Ces 4.33 mL of 0.1274 M KIO3 La was required. Calculate the mass fraction of La and Ce in the sample. Number Number g La g Ce 0.337 0.663 g sample g sample Incorrect. Previous 3 Give Up & View Solution Try Again Next Exit Explanation

Explanation / Answer

1 mole LaCl3 reacts with 3 moles KIO3

1 mole Ce(NO3)3 also reacts with 3 moles of KIO3

moles KIO3 = 0.1274 M x 44.33 ml = 5.65 mmol

moles of LaCl3 = moles of Ce(NO3)3 = 5.65/2 x 3 = 0.942 mmol

mass fraction La in the sample = 0.942 x 138.9/0.5087 = 0.257

mass fraction Ce in the sample = 0.942 x 140.1/0.5087 = 0.177

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