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1 year ago

(ii) Calculate the maximum mass of ethanol that could be obtained g from 30.0 g of glucose.

Chemistry

1 answer:

Step2247 [10]1 year ago

3 0

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

<h3>Mass of C₂H₅OH formed</h3>

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?

$mass of ethanol=\frac{30 grams of glucosex 92 grams of ethanol}{180 grams of glucose}$

<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 fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

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$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

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