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

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

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Answer 1 · 2017-04-20T12:25:51

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

........and we get approx. $29 \cdot k g$ $29*kg$ of carbon dioxide.

Explanation:

We will be requiring $(i)$ $(i)$ $C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$ $C_6H_6(l) + 15/2O_2(g) rarr 6CO_2(g) + 3H_2O(l)$

$(i i)$ $(ii)$ the density of benzene, $ρ_{benzene} = 0.88 \cdot g \cdot m L^{- 1}$ $rho_"benzene"=0.88*g*mL^-1$ ,

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

$Moles of benzene = \frac{9.75 \cdot L \times 0.88 \cdot g \cdot m L^{- 1} \cdot 10^{3} \cdot m L \cdot L^{- 1}}{78.11 \cdot g \cdot m o l^{- 1}}$ $"Moles of benzene"=(9.75*Lxx0.88*g*mL^-1*10^3*mL*L^-1)/(78.11*g*mol^-1)$

$= 109.9 \cdot m o l$ $=109.9*mol$ .

Given the stoichiometry, the resultant mass of carbon dioxide is...... $6 \times 109.9 \cdot m o l \times 44.01 \cdot g \cdot m o l^{- 1} ≅ 29 \cdot k g$ $6xx109.9*molxx44.01*g*mol^-1~=29*kg$

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

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

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