Asked by robie miranda
3. A carbonate fusion was needed to free the Bi from a 0.6423-g sample of eulytite (2Bi2O3ยท3SiO2). The fused mass was dissolved in dilute acid, following which the Bi+3 was titrated with 27.36mL of 0.03369M NaH2PO4. The reaction is Bi+3 + H2PO4- ๏ BiPO4(s) + 2H+. Calculate the percent purity of eulytite (1112g/mol) in the sample?
Answers
Answered by
DrBob222
mols NaH2PO4 = M x L = ?
mols Bi^3+ = mols NaH2PO4 from the coefficients in the balanced titration equation (1:1).
Convert mols Bi^3+ to mols eulytite by mols Bi^3+ x (molar mass eulytite/2*atomic mass Bi) = ?
Then g eulytite = mols eulytite x molar mass eulytite.
Then % eulytite = (g/mass sample)*100 = ?
mols Bi^3+ = mols NaH2PO4 from the coefficients in the balanced titration equation (1:1).
Convert mols Bi^3+ to mols eulytite by mols Bi^3+ x (molar mass eulytite/2*atomic mass Bi) = ?
Then g eulytite = mols eulytite x molar mass eulytite.
Then % eulytite = (g/mass sample)*100 = ?
Answered by
Alice
this solution was false
Answered by
Valens
The answer given to this question in most of the online answers is not correct, the correct answer is ๐๐.๐%
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