A #17.8L# volume of propane gas for which #rho=493g*L^-1# was completely combusted. What are (i) the molar quantities, and (ii) the mass of carbon dioxide emitted?

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Answer 1 · 2016-12-19T13:27:07

Approx. #26.4*kg*CO_2# are emitted.

We need (i) a stoichiometrically balanced equation:

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(l)#

And (ii) the equivalent quantity of propane gas.

#"Moles of propane"# #=# #(17.8*cancelLxx493*cancelg*cancel(L^-1))/(44.1*cancelg*mol^-1)=199*mol#

Now clearly, almost #600*mol# of carbon dioxide will be evolved. Why clearly?

And thus mass of #"carbon dioxide"# emitted,

#600*molxx44.01*g*mol^-1~=9*kg*CO_2#

A #17.8*L# volume of propane gas for which #rho=493*g*L^-1# was completely combusted. What are (i) the molar quantities, and (ii) the mass of carbon dioxide emitted?