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

Question

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?

Answer 1

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?

Answer 2

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 = ?

Answer 3

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.