The substance ethanol (c2h6o) is readily combustedin air, i. e. reacts with oxygen, to form water andcarbon dioxide. if 125 g of ethanol is burnedcompletely, how much oxygen is consumed? if 127 g of water is recovered in the reaction, whatis the yield for water? how much carbon dioxide isexpected to be formed? if 221 g is isolated, what isthe yield?

a) 260.16 g of O2

b) 86.78% of water

c) 238.48 g

d) 92.67%

Explanation:

First, we need to write the reaction that is taking place and balance if it's neccesary:

C2H6O + 3O2 > 2CO2 + 3H2O

Now that we have the reaction balanced, we can solve the problem for parts.

a) oxygen consumed

in this part, we need to know how much of oxygen reacted with the 125 g of ethanol. We can know this, calculating the moles of ethanol, and then, do a stechiometry with the oxygen.

For the moles of ethanol we need the molar mass of it, which is 46.07 g/mol so the moles:

moles ethanol = 125 / 46.07 = 2.71 moles

now for stechiometry, we know that 1 mole of ethanol reacts with 3 moles of oxygen, so doing a math of 3:

moles of Oxygen = 2.71 * 3 = 8.13 moles

for the mass of oxygen, we need the molar mass of oxygen which is 32 g/mol so:

mO2 = 32 * 8.13 = 260.16 g of O2

b) yield of water

To know this, we need to calculate the theorical mass of water needed.

Assuming we have the same grams of ethanol from part a), and that it has the same relation of moles with oxygen (3 moles of oxygen produces 3 moles of water), the moles of water are 8.13 moles, so it's mass:

mH2O = 18 * 8.13 = 146.34 g

This is the theorical yield, so the actual yield of water would be:

% = 127 / 146.34 * 100 = 86.78% of water

c) We'll do the same procedure of before, but with CO2, so:

moles of CO2 = 2.71 * 2 = 5.42 moles

mass of CO2 = 5.42 * 44 = 238.48 g

This is the mass of expected CO2 to be formed, the theorical yield

d) the yield for CO2 knowing that it was isolated 221 g.

% = 221 / 238.48 * 100

% = 92.67% of CO2

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole