Asked by Colby
The tin (II) fluoride in a 10.000 g sample of toothpaste is extracted into solution and them a precipitate of lanthanum (II) fluoride (LaF3) is formed by the addition of a lanthanum (III) nitrate solution. If the mass of the lanthanum (III) fluoride precipitate is 0.105g, what is the percentage of tin (II) fluoride in the toothpaste?
Answers
Answered by
DrBob222
3SnF2 + 2La(NO3)3 ==> 2LaF3 + 3Sn(NO3)2
1.mols SnF2 = grams/molar mass
2.Using the coefficients in the balanced equation, convert mols SnF2 to mols LaF3.
3. Convert mols of LaF3 to grams. g = mols x molar mass. This is the theoretical yield (TY) of LaF3 from the reaction. The actual yield (AY) is 0.105 g.
4. %yield = (AY/TY)*100 = ?
1.mols SnF2 = grams/molar mass
2.Using the coefficients in the balanced equation, convert mols SnF2 to mols LaF3.
3. Convert mols of LaF3 to grams. g = mols x molar mass. This is the theoretical yield (TY) of LaF3 from the reaction. The actual yield (AY) is 0.105 g.
4. %yield = (AY/TY)*100 = ?
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