C3H8(g)+5O2(g)→ 3CO2(g)+4H2O(l)

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Answer #1

So we gots...#C_3H_8 + 5O_2(g) rarr 3CO_2(g) +4H_2O(l)#


Clearly, four moles of water result from each mole of propane.

And if #1.638*g# of propane are combusted this represents a molar quantity of...

#(1.638*g)/(44.10*g*mol^-1)=0.0371*mol#...with respect to propane.

And thus .....

#"mass"_"carbon dioxide"=3xx0.0371*molxx44.01*g*mol^-1=4.90*g#


How does the total mass of the products compare to the total mass of the reactants? Is this significant conceptually?

answered by: anor277
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( a) In the complete combustion of propane, how many moles of H2O(l) are produced per mole of CO2(g)? (b) A 1.638-g sample of propane is burned in excess oxygen. What are the theoretical yields (in grams) of CO2(g) and H2O(l) expected from the reaction
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