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?
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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.
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
To answer this question, we need to calculate the stoichiometry of the reaction and use it to determine the minimum amount of toluene needed to achieve a yield of 71% and produce 11.5 grams of potassium benzoate (KC7H5O2).
First, let's calculate the molar mass of toluene (C7H8) and potassium benzoate (KC7H5O2):
- Molar mass of toluene (C7H8) = (7 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
= (7 * 12.01 g/mol) + (8 * 1.01 g/mol) = 92.14 g/mol
- Molar mass of potassium benzoate (KC7H5O2) = atomic mass of potassium + (7 * atomic mass of carbon) + (5 * atomic mass of hydrogen) + (2 * atomic mass of oxygen)
= 39.10 g/mol + (7 * 12.01 g/mol) + (5 * 1.01 g/mol) + (2 * 16.00 g/mol) = 160.21 g/mol
Now, let's calculate the amount of potassium benzoate (KC7H5O2) produced from 11.5 grams, considering a 71% yield:
- Amount of potassium benzoate (KC7H5O2) = (11.5 g / 100) * (100 / 71%) = 16.20 g
Next, we need to calculate the molar ratio between toluene (C7H8) and potassium benzoate (KC7H5O2) from the balanced chemical equation:
- Balanced equation: C7H8 + 2 KMnO4 --> 2 MnO2 + KOH + H2O
- The ratio between toluene (C7H8) and potassium benzoate (KC7H5O2) is 1:1.
Now, we can calculate the minimum amount of toluene needed to achieve a yield of 71% and produce 11.5 grams of potassium benzoate (KC7H5O2):
- Moles of potassium benzoate = mass / molar mass = 16.20 g / 160.21 g/mol = 0.101 moles
- Since the ratio between toluene and potassium benzoate is 1:1, we need 0.101 moles of toluene to get the same amount of potassium benzoate.
- Moles of toluene = 0.101 moles
- Mass of toluene = moles * molar mass = 0.101 moles * 92.14 g/mol = 9.32 grams (approximately)
Therefore, the minimum number of grams of toluene needed to achieve a yield of 71% and produce 11.5 grams of potassium benzoate is approximately 9.32 grams.