Respuesta :
Answer:
0.708 L
Explanation:
Start with the balanced chemical equation:
[tex]2C_{8}H_{18} +25O_{2} --->16CO_{2} +18H_{2}O[/tex]
calculate the number of moles of octane:
n(octane)=[tex]\frac{mass}{molar mass}=\frac{0.210g}{114.23g/mol}=1.8384*10^{-3}mol[/tex]
From the balanced equation we see that the number of carbon dioxide produced is always 16 times the number of moles of octane available.
Therefore n(carbon dioxide)=[tex]1.8384*10^{-3} mol *16=0.0294144mol[/tex]
We now treat CO2 as an idal gas, which in reality it behaves like. That allows us to use the following ideal gas equation.
[tex]Volume=\frac{nRT}{P}=\frac{0.0294144mol*0.08205L.atm/(mol.K)*293.15K}{1 atm} =0.7075L[/tex]
Answer: The volume of carbon dioxide gas that is produced from the given amount of octane is [tex]3.29\times 10^2L[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of octane = 0.210 kg = 210 g (Conversion factor: 1 kg = 1000 g)
Molar mass of octane = 114.23 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of octane}=\frac{210g}{114.23g/mol}=1.84mol[/tex]
The temperature and pressure conditions given to us are NTP conditions:
At NTP:
1 mole of a gas occupies 22.4 L of volume.
So, 1.84 moles of octane gas will occupy [tex]22.4\times 1.84=41.216L[/tex] of volume.
The chemical reaction for the combustion of octane follows the equation:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
By Stoichiometry of the reaction:
[tex](2\times 22.4L)[/tex] of octane gas produces [tex](16\times 22.4L)[/tex] of carbon dioxide gas.
So, 41.216 L of octane gas will produce = [tex]\frac{(16\times 22.4)L}{(2\times 22.4)L}\times 41.216=329.728L=3.29\times 10^2L[/tex] of carbon dioxide gas.
Hence, the volume of carbon dioxide gas that is produced from the given amount of octane is [tex]3.29\times 10^2L[/tex]
