A car burns octane in the following balanced reaction: 2C8H18 + 25O2 -> 16CO2 + 18H2O. If the car burns 500 grams of octane, how many grams of CO2 will be produced?

The answer to your question is 1544 g of CO₂

Explanation:

Data

mass of octane = 500 g

mass of CO₂ = ?

Balanced chemical reaction

               2C₈H₁₈  +  25O₂   ⇒    16CO₂   +   18H₂O

Process

1.- Calculate the molar mass of octane and of carbon dioxide

C₈H₁₈ = 2[(12 x 8) + (1 x 18)] = 2[(96 + 18] = 2[114] = 228 g

CO₂ = 16[(12 x 1) + (16 x 2)] = 16[12 + 32] = 16[44] = 704 g

2.- Calculate the mass of carbon dioxide produced

                    228 g of C₈H₁₈ 704 g of CO₂

                     500 g of C₈H₁₈  x CO₂

                              x = (500 x 704) / 228

                              x = 352000/228

                              x = 1544 g of CO₂

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%