Question

In a laboratory, 23.5 g of cyclohexane is burned and 5.5 liters of carbon dioxide was obtained ? a) Balance the reaction b) Calculate the reaction yield. d(CO2)= 1.9 g/L

Answer 1

`"Theoretical Yield"=73.86g`
`"Actual Yield"=10.45g`

Explanation:

1. Write and balance the equation
   `C_6H_12+9O_2->6CO_2+6H_2O`
2. Find the molar masses of the involved compounds that can be used later for the usual molar conversions.
   `C_6H_12=(84g)/(mol)`
   `CO_2=(44g)/(mol)`
3. Given the mass of `C_6H_12`, per convention, convert it to mole (`eta`). Knowing the fact as shown above that `1molC_6H_12-=84gC_6H_12`, a conversion factor is obtainable from this relationship; i.e.,
   `=23.5cancel(gC_6H_12)xx(1molC_6H_12)/(84cancel(gC_6H_12))`
   `=0.2798molC_6H_12`
4. Then, find the mole of `CO_2`. Given the relationship `1molC_6H_12-=6molCO_2` from the balanced equation, a factor used for the conversion is obtainable; i.e.,
   `=0.2798cancel(molC_6H_12)xx(6molCO_2)/(1cancel(molC_6H_12))`
   `=1.6786molCO_2`
5. Now, find the Theoretical Yield `(TY)` of this reaction.
   `TY=1.6786cancel(molCO_2)xx(44gCO_2)/(1cancel(molCO_2))`
   `color(red)(TY=73.86gCO_2`
6. Given the volume produced in the lab and the density of the `CO_2`, the Actual Yield `(AY)` can be computed as:
   `rho=m/V`
   `m=rhoxxV`
   `m=(1.9g)/cancel((L))xx5.5cancel(L)`
   `color(blue)(m=10.45gCO_2=AY`