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

Answer 1

Cameron Andrews

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#

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Answer 2

Noah Aldrich

The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L. For carbon dioxide, the molar mass is approximately 44 g/mol. Therefore, the mass of carbon dioxide from the complete combustion of a 9.75 L volume of benzene is approximately ( \frac{9.75 , \text{L}}{22.4 , \text{L/mol}} \times 44 , \text{g/mol} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.