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Sagot :
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:
[tex]\begin{gathered} PV=\text{nRT,} \\ n=\frac{PV}{RT}, \\ n=\frac{\text{1 atm }\cdot35.5\text{ L}}{0.082\frac{atm\cdot L}{mol\cdot K}\cdot273K}, \\ n(C_5H_{12})=1.58\text{6 moles} \end{gathered}[/tex]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:
[tex]1.586molC_5H_{12}\cdot\frac{5molCO_2}{1molC_5H_{12}_{}}=7.93molCO_2.[/tex]The next and final step is clear V (volume) from the initial formula and replaces the value of moles for CO2, like this:
[tex]\begin{gathered} V=\frac{nRT}{P}, \\ V=\frac{7.93\text{ mol}\cdot0.082\frac{atm\cdot L}{mol\cdot K}\cdot273K}{1\text{ atm}}, \\ V(CO_2)=177.52\text{ L.} \end{gathered}[/tex]So, 25,5 L of C5H12 will produce 177.52L of CO2.
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