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

Dec 17, 2017

$\text{Methane}$ is $C {H}_{4}$; $\text{propane}$ is ${C}_{3} {H}_{8}$... 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.

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + \Delta$

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

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right) + \Delta$

$\text{Moles of propane} \equiv \frac{295 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} = 6.69 \cdot m o l$...

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 $22.4 \cdot L \cdot m o {l}^{-} 1$...we get...

$3 \times 6.69 \cdot m o l \times 22.4 \cdot L \cdot m o {l}^{-} 1 \cong 450 \cdot L$....take that environment...

Dec 17, 2017

450.55L $C {O}_{2}$ produced

Explanation:

!!! ${C}_{3}$${H}_{8}$ is PROPANE, not methane

The chemical equation for the combustion of propane:
${C}_{3}$${H}_{8}$ + $5 {O}_{2}$ -> $3 C {O}_{2}$ + $4 {H}_{2}$$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}$${H}_{8}$)
* ( 3mol $C {O}_{2}$ $/$ 1mol ${C}_{3}$${H}_{8}$) * ( 22.4L $C {O}_{2}$ $/$ 1mol $C {O}_{2}$) --->

450.55L $C {O}_{2}$ produced