Answer:
1. 79.2 g of CO2
2. the reaction is spontaneous and favors the formation of products
3. CH2O
Step-by-step explanation:
To solve the problem, we first need to calculate the limiting reagent by comparing the amount of glucose and oxygen available for the reaction. We will assume that the reaction goes to completion.
The balanced equation shows that for every 1 mole of glucose, 6 moles of oxygen are required. Therefore, the moles of oxygen required for 9.0 grams of glucose is:
moles of glucose = mass/molar mass = 9.0/180.16 = 0.0499 mol
moles of oxygen = 6 x moles of glucose = 6 x 0.0499 = 0.2994 mol
Since we have 0.2994 moles of oxygen available, and only 0.2000 moles of oxygen are required to react with 0.0499 moles of glucose to produce 0.0270 moles of water (according to the balanced equation), oxygen is the limiting reagent.
Using the balanced equation, we can now calculate the moles of CO2 produced:
moles of water produced = mass/molar mass = 5.4/18.02 = 0.2997 mol
moles of CO2 produced = 6 x moles of water produced = 6 x 0.2997 = 1.7982 mol
Finally, we can calculate the mass of CO2 produced:
mass of CO2 produced = moles x molar mass = 1.7982 x 44.01 = 79.2 g
Therefore, the mass of CO2 produced when 9.0 grams of glucose completely reacts with 9.6 grams of oxygen to produce 5.4 grams of water is 79.2 g.
To compare the entropy of the reactants to the entropy of the products, we can use the equation:
ΔS = ΣS(products) - ΣS(reactants)
The entropy of a substance depends on its state and temperature, and can be looked up in tables. At standard conditions (298 K and 101.3 kPa), the molar entropy of glucose, oxygen, CO2, and liquid water are:
S(C6H12O6) = 212.8 J/(mol K)
S(O2) = 205.0 J/(mol K)
S(CO2) = 214.8 J/(mol K)
S(H2O) = 69.9 J/(mol K)
Using the above values and the balanced equation, we can calculate the entropy change:
ΔS = (6 x S(CO2) + 6 x S(H2O)) - (S(C6H12O6) + 6 x S(O2))
ΔS = (6 x 214.8 + 6 x 69.9) - (212.8 + 6 x 205.0)
ΔS = 287.4 J/(mol K)
Since ΔS is positive, the entropy of the products is greater than the entropy of the reactants. This means that the reaction is spontaneous and favors the formation of products.
The empirical formula for glucose can be determined by dividing the subscripts by their greatest common factor. In this case, the empirical formula is:
C6H12O6 ÷ 6 = CH2O
Therefore, the empirical formula for glucose is CH2O.