Question

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

Answer 1

`"Methane"` is `CH_4`; `"propane"` is `C_3H_8`... for propane we get....

Explanation:

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...

Answer 2

450.55L `CO_2` produced

Explanation:

!!! `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