Under standard conditions, what volume of carbon dioxide would result from the combustion of #295*g# of #"propane"#?

We need to write a stoichiometrically balanced equation. And for combustion reactions the usually rigmarole is to (i) balance the carbons as carbon dioxide; (ii) then balance the hydrogens as water, and (iii) then balance the oxygens.

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

Even-numbered alkanes require a half-integral coefficient for dioxygen....

#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) +5H_2O(l)+Delta#

#"Moles of propane"-=(295*g)/(44.1*g*mol^-1)=6.69*mol#...

And thus, upon complete combustion, we gets #3xx6.69*molxx44.01*g*mol^-1=??*g# with respect to carbon dioxide.

At STP, if the molar volume is #22.4*L*mol^-1#...we get...

#3xx6.69*molxx22.4*L*mol^-1~=450*L#....take that environment...

!!! #C_3##H_8# is PROPANE, not methane

The chemical equation for the combustion of propane:
#C_3##H_8# + #5O_2# -> #3CO_2# + #4H_2##O#

Since there's an excess of oygen, propane gas is the limiting reactant. Using dimensional analysis, find how many liters of carbon dioxide are produced:

(1mol = 22.4L @ STP)

( 295g #C_3##H_8#) * ( 1mol #C_3##H_8# #//# 44g #C_3##H_8#)
* ( 3mol #CO_2# #//# 1mol #C_3##H_8#) * ( 22.4L #CO_2# #//# 1mol #CO_2#) --->

450.55L #CO_2# produced