Combustion of which gas gives 7 moles of gaseous product from 6 moles of gaseous reactant?

#1.# #CH_4;#
#2.# #H_2C=CH_2;#
#3.# #HC-=CH;#
#4.# #H_3C-CH_2CH_2#

Answer 1

Jeremiah Adams

Propane combustion; #"option 4"#

#underbrace(C_3H_8(g) + 5O_2(g))_"6 moles" rarr underbrace(3CO_2(g) +4 H_2O(g))_"7 moles"#

Because #Vpropn" (T and P constant)"#, the VOLUMES represent the stoichiometric equivalents of hydrocarbon and oxidant.

You should write out all the combustion reactions with each fuel to satisfy yourself that this is the case:

i.e. #C_2H_4(g) +3O_2(g) rarr 2CO_2(g) +2H_2O(g)#, which stoichiometry disagrees with the experimental result.

You can do the rest.

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Answer 2

Luna Chen

Six moles of gaseous reactant and seven moles of gaseous product are produced during the combustion of propane gas (C3H8).

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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.

Combustion of which gas gives 7 moles of gaseous product from 6 moles of gaseous reactant?

#1.# #CH_4;# #2.# #H_2C=CH_2;# #3.# #HC-=CH;# #4.# #H_3C-CH_2CH_2#

#1.# #CH_4;#
#2.# #H_2C=CH_2;#
#3.# #HC-=CH;#
#4.# #H_3C-CH_2CH_2#