Asked by Veronica
The potassium salt of benzoic acid, potassium benzoate, KC7H5O2, can be made by the action of potassium permanganate on toluene, C7H8.
C7H8 + 2 KMnO4 --> 2MnO2 + KOH + H2O
If the yield of potassium benzoate cannot be realistically be expected to be more than 71%, what is the minimum number of grams of toluene needed to achieve this yield while producing 11.5 g of potassium benzoate?
C7H8 + 2 KMnO4 --> 2MnO2 + KOH + H2O
If the yield of potassium benzoate cannot be realistically be expected to be more than 71%, what is the minimum number of grams of toluene needed to achieve this yield while producing 11.5 g of potassium benzoate?
Answers
Answered by
Veronica
Nevermind I got the answer myself!
Answered by
DrBob222
You had a problem almost like this that Bob Pursley helped you with.
You ant 11.5 g pot benzoate so the reaction must produce a theoretical yield of 11.5g/0.71 = ?
Then mols K benzoate = ?/molar mass K benzoate
Convert to mols tolune, then
g tolune = mols x molar mass toluene.
You ant 11.5 g pot benzoate so the reaction must produce a theoretical yield of 11.5g/0.71 = ?
Then mols K benzoate = ?/molar mass K benzoate
Convert to mols tolune, then
g tolune = mols x molar mass toluene.
Answered by
Ronn
C7H8 + 2KMnO4 ---------- KC7H5O2 + 2MnO2 + KOH + H2O
If the yield of potassium benzoate cannot realistically be expected to be more than 71%, what is the minimum number of grams of toluene needed to produce 11.5 g of potassium benzoate?
Group of answer choices
9.2 g
16 g
8.02 g
22.6 g
If the yield of potassium benzoate cannot realistically be expected to be more than 71%, what is the minimum number of grams of toluene needed to produce 11.5 g of potassium benzoate?
Group of answer choices
9.2 g
16 g
8.02 g
22.6 g
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