3. hexane (c6h-2) is combusted in oxygen
balanced reaction:
a. how many moles of oxygen are needed to react with 1 mole of hexane?
b. how many molecules of carbon dioxide can be produced from 245.5
grams of hexane?

                   a)  9.5 moles of Oxygen

                    b)  1.029 × 10²⁵ Molecules

Explanation:

                    The balance chemical equation for said reaction is as follow:

                               2 C₆H₁₄ + 19 O₂   →  12 CO₂ + 14 H₂O

A)

According to equation,

              2 moles of C₆H₁₄ are reacted with  =  19 moles of O₂

So,

                   1 moles of C₆H₁₄ will react with  =  X moles of O₂

Solving for X,

                      X  =  19 moles × 1 mole / 2 moles

                       X  =  9.5 moles of Oxygen

B)

For this part first calculate moles of hexane as;

                             Moles  =  Mass / M.Mass

                             Moles  =  245.5 g / 86.18 g/mol

                             Moles  =  2.84 mole of Hexane

Secondly, find the number of moles of CO₂ as,

According to equation,

              2 moles of C₆H₁₄ produced  =  12 moles of CO₂

So,

                   2.84 moles of C₆H₁₄ will produce  =  X moles of CO₂

Solving for X,

                      X  =  12 moles × 2.84 mole / 2 moles

                       X  =  17.09 moles of CO₂

Lastly, calculate the number of molecules of CO₂ as,

Number of Molecules  =  Moles × 6.022 × 10²³ molecules/mole

Number of Molecules  =  17.09 moles × 6.022 × 10²³ molecules/mole

Number of Molecules  =  1.029 × 10²⁵ Molecules

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