What mass of carbon dioxide results from complete combustion of a $9.75*L$ volume of benzene?

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Answer 1 · 2017-04-20T12:25:51

We need (i) a stoichiometric equation...........

........and we get approx. $29*kg$ of carbon dioxide.

We will be requiring $(i)$ $C_6H_6(l) + 15/2O_2(g) rarr 6CO_2(g) + 3H_2O(l)$

$(ii)$ the density of benzene, $rho_"benzene"=0.88*g*mL^-1$ ,

And $(iii)$ , equivalent quantities of benzene........

$"Moles of benzene"=(9.75*Lxx0.88*g*mL^-1*10^3*mL*L^-1)/(78.11*g*mol^-1)$

$=109.9*mol$ .

Given the stoichiometry, the resultant mass of carbon dioxide is...... $6xx109.9*molxx44.01*g*mol^-1~=29*kg$

What mass of carbon dioxide results from complete combustion of a 9.75*L9.75*L volume of benzene?