Question

REACTION; C5H12 + 8O2 5CO2 + 6H2OWhen 35.5 L of C5H12 are consumed in this reaction what volume of CO2 can be produced in liters?

Answer 1

Using the STP (standard temperature and pressure), we can solve this problem. First, let's find the number of moles produce by 35.5 L of C5H12.

Remember that the standard temperature is 0 °C which is the same that 273 K (kelvin) and for pressure is 1 atm and the constant of ideal gas is 0.082 atm*L/(mol*K)

Let's use the formula of an ideal gas to find the number of moles:

`\begin{gathered} PV=\text{nRT,} \\ n=(PV)/(RT), \\ n=\frac{\text{1 atm }\cdot35.5\text{ L}}{0.082(atm\cdot L)/(mol\cdot K)\cdot273K}, \\ n(C_5H_(12))=1.58\text{6 moles} \end{gathered}`

Now, using this number of moles, we can find the number of moles produce for CO2 and we can find its volume.

You can see that in the reaction 1 mol of C5H12 produces 5 moles of CO2, so the calculation to find the number of moles of CO2, would be:

`1.586molC_5H_(12)\cdot\frac{5molCO_2}{1molC_5H_(12)_{}}=7.93molCO_2.`

The next and final step is clear V (volume) from the initial formula and replaces the value of moles for CO2, like this:

`\begin{gathered} V=(nRT)/(P), \\ V=\frac{7.93\text{ mol}\cdot0.082(atm\cdot L)/(mol\cdot K)\cdot273K}{1\text{ atm}}, \\ V(CO_2)=177.52\text{ L.} \end{gathered}`

So, 25,5 L of C5H12 will produce 177.52L of CO2.