The fermentation of glucose to form ethanol occurs according to the following chemical

reaction:
C6H12O6(aq) → 2 C2H5OH(l) + 2 CO2(g)
A) If 700 grams of glucose (C6H12O6) are fermented, what is the maximum volume in milliliters of ethanol (C2H5OH) that can form? (Density of ethanol is 0.79 g/mL)
B)2 kg of glucose is fermented and 1 L of ethanol is formed. What is the percent yield of this reaction?

Calcium Sulfate (CaSO4) decomposes at high temperatures according to the reaction:

2 CaSO4(s) → 2 CaO(s) + 2 SO2(g) + O2(g)
- If CaSO4 is sealed in a chamber, heated to a high temperature, and allowed to proceed to equilibrium. Then, the pressure of the gaseous equilibrium products are evacuated , lowering the pressure of the system.
A) Will this shift equilibrium towards the reactants, products, or have no effect on
equilibrium?
B) Will this increase, decrease, or not change the amount of CaSO4 present?
C) Will this increase, decrease, or not change the amount of CaO present?

Please don't post another question as an answer to another question.

A. See below for ethanol problem.
B. Convert 2 kg glucose to L ethanol formed using the same procedure used in the previous problem. This will give you the theoretical yield (TY)( that's the yield as if it were 100%). The actual yield (AY) in the problem is listed as 1 L.
%yield = (AY/TY)*100 = ?

You are removing products. Le Chatelier's Principle says that if a system in equilibrium is disturbed, then the system will shift so as to undo what we did to it. Therefore, it we remove the products, more CaSO4 will decompose to form the products we removed. You can take it from here.

A) To find the maximum volume of ethanol that can form, we need to use the information provided along with the stoichiometry of the reaction.

1 mole of glucose (C6H12O6) produces 2 moles of ethanol (C2H5OH). The molecular weight of glucose is 180 g/mol, and the molecular weight of ethanol is 46 g/mol.

First, we need to calculate the number of moles of glucose in 700 grams:
Number of moles of glucose = mass of glucose / molar mass of glucose
Number of moles of glucose = 700 g / 180 g/mol
Number of moles of glucose = 3.889 moles

Now, based on the stoichiometry of the reaction, we know that 1 mole of glucose produces 2 moles of ethanol. Therefore, the number of moles of ethanol formed will be twice the number of moles of glucose:

Number of moles of ethanol = 2 x Number of moles of glucose
Number of moles of ethanol = 2 x 3.889 moles
Number of moles of ethanol = 7.778 moles

Next, we can use the density of ethanol to calculate the volume:
Volume of ethanol = mass of ethanol / density of ethanol
Volume of ethanol = (molar mass of ethanol x number of moles of ethanol) / density of ethanol
Volume of ethanol = (46 g/mol x 7.778 moles) / 0.79 g/mL
Volume of ethanol = 450.663 mL

Therefore, the maximum volume of ethanol that can form from 700 grams of glucose is approximately 450.663 mL.

B) To calculate the percent yield of the reaction, we need to compare the actual yield (1 L) with the theoretical yield.

First, we need to calculate the theoretical yield of ethanol from 2 kg of glucose. We already know that 1 mole of glucose produces 2 moles of ethanol. The molar mass of glucose is 180 g/mol, so:

Number of moles of glucose = mass of glucose / molar mass of glucose
Number of moles of glucose = 2000 g / 180 g/mol
Number of moles of glucose = 11.111 moles

Theoretical yield of ethanol = 2 x Number of moles of glucose
Theoretical yield of ethanol = 2 x 11.111 moles
Theoretical yield of ethanol = 22.222 moles

Now, we need to convert the theoretical yield from moles to liters:
Theoretical yield in liters = Theoretical yield in moles x 22.4 L/mol (molar volume at STP)
Theoretical yield in liters = 22.222 moles x 22.4 L/mol
Theoretical yield in liters = 497.774 L

Finally, we can calculate the percent yield of the reaction:
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (1 L / 497.774 L) x 100
Percent yield = 0.201%

Therefore, the percent yield of the reaction is approximately 0.201%.