A 0.52900.5290 g solid sample containing a mixture of LaCl3LaCl3 (molar mass = 245.26 g/mol) and...
A 0.52900.5290 g solid sample containing a mixture of LaCl3LaCl3 (molar mass = 245.26 g/mol) and Ce(NO3)3Ce(NO3)3 (molar mass = 326.13 g/mol) was dissolved in water. The solution was titrated with KIO3KIO3, producing the precipitates La(IO3)3La(IO3)3 and Ce(IO3)3Ce(IO3)3. For the complete titration of both La3+La3+ and Ce3+Ce3+, 42.2742.27 mL of 0.12130.1213 M KIO3KIO3 was required. Calculate the mass fraction of LaLa and CeCe in the sample.
mass fraction La:______ g LA/ g Sample
mass fraction Ce:______ g Ce/ g Sample
Homework Answers
First mass of LaCl3 is calculated , then moles of LaCl3 is same as moles of La. For calculations see the picture below:
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A 0.52900.5290 g solid sample containing a mixture of LaCl3LaCl3
(molar mass = 245.26 g/mol) and...
A 0.5137 g solid sample containing a mixture of LaCl, (molar mass = 245.26 g/mol) and...
A 0.5137 g solid sample containing a mixture of LaCl, (molar mass = 245.26 g/mol) and Ce(NO3), (molar mass = 326.13 g/mol) was dissolved in water. The solution was titrated with KIO, producing the precipitates La(IO,), and Ce(IO). For the complete titration of both La and Ce+, 42.25 mL of 0.1204 M KIO, was required. Calculate the mass fraction of La and Ce in the sample. mass fraction La: g La g sample mass fraction Ce: g Ce g sample
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A 49.5 ml solution containing 3.03 g of Na2CO3 (molar mass = 106 g/mol) and 50.5...
A 49.5 ml solution containing 3.03 g of Na2CO3 (molar mass = 106 g/mol) and 50.5 ml solution containing excess Al(NO3)3 (molar mass = 213 g/mol) are mixed. What is the molecular formula of the insoluble product and how much of that product is formed? Al2(CO3)3; 2.23 g formed AICO; 1133 g formed Al2(CO3)3: 6.38 g formed Nag(NO3)63.03 g formed NaNO3;2.23 g formed NaNO3: 6.38 g formed
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A 0.317 g sample of a monoprotic acid is dissolved in water and titrated with 0.100 M KOH. What is the molar mass of th...
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A 0.258-g sample of a pure triprotic acid, H3A,
(where A is the generic anion of the acid), was dissolved in water
and titrated with 0.150 M barium hydroxide solution. The titration
required 13.9 mL of the base to reach the equivalence point. What
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