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

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Answer 1 · 2017-12-23T01:59:24

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

Write and balance the equation
$C_6H_12+9O_2->6CO_2+6H_2O$
Find the molar masses of the involved compounds that can be used later for the usual molar conversions.
$C_6H_12=(84g)/(mol)$
$CO_2=(44g)/(mol)$
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.,
$=23.5cancel(gC_6H_12)xx(1molC_6H_12)/(84cancel(gC_6H_12))$
$=0.2798molC_6H_12$
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.,
$=0.2798cancel(molC_6H_12)xx(6molCO_2)/(1cancel(molC_6H_12))$
$=1.6786molCO_2$
Now, find the Theoretical Yield $(TY)$ of this reaction.
$TY=1.6786cancel(molCO_2)xx(44gCO_2)/(1cancel(molCO_2))$
$color(red)(TY=73.86gCO_2$
Given the volume produced in the lab and the density of the $CO_2$ , the Actual Yield $(AY)$ can be computed as:
$rho=m/V$
$m=rhoxxV$
$m=(1.9g)/cancel((L))xx5.5cancel(L)$
$color(blue)(m=10.45gCO_2=AY$