(ii) Calculate the maximum mass of ethanol that could be obtained g from 30.0 g of glucose.
1 answer:
Taking into account the reaction stoichiometry, 15.33 grams ethanol could be obtained from 30 grams of glucose.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- C₆H₁₂O₆: 1 mole
- C₂H₅OH: 2 moles
- CO₂: 2 moles
The molar mass of the compounds is:
- C₆H₁₂O₆: 180 g/mole
- C₂H₅OH: 46 g/mole
- CO₂: 44 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₆H₁₂O₆: 1 mole ×180 g/mole= 180 grams
- C₂H₅OH: 2 moles ×46 g/mole= 92 grams
- CO₂: 2 moles ×44 g/mole= 88 grams
The following rule of three can be applied: if by reaction stoichiometry 180 grams of glucose form 92 grams of ethanol, 30 grams of glucose form how much mass of ethanol?
<u><em>mass of ethanol= 15.33 grams</em></u>
Then, 15.33 grams ethanol could be obtained from 30 grams of glucose.
Learn more about the reaction stoichiometry:
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The answer to the question is a.
Answer:
0.17%
Explanation:
With the equation:
2Cr2O7 2- + C2H5OH + H2O --> 4Cr3+ + 2CO2 + 11H2O
We can assume that every mole of ethanol needs 2 moles of Dichromate to react.
So if in 1L we have 0.05961 moles of dichromate we can discover how many moles we have in 35.46mL
1000 mL - 0.05962 moles
35.46 mL - x
x =
x = 2,11* 10^-3 moles
As we said earlier, 1 mole of ethanol needs 2 mole of dichromate, so in the solution we have 1,055*10^-3 moles of ethanol. We can discover the mass of ethanol present in the solution.
1 mole - 46g
1.055*10^-3 - y
y = 46 * 1.055*10^-3
y = 0.048 g
To discover the percent of alchol we can use a simple relation
28 g - 100%
0.048 - z
z =
z = 0.17%
Answer:
here is definition with example
