Answer:
The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}
Explanation:
1. Write down the balanced chemical reaction:
![2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}](https://tex.z-dn.net/?f=2C_%7B6%7DH_%7B14%7D_%7B%28l%29%7D%2B19O_%7B2%7D_%7B%28g%29%7D%3D12CO_%7B2%7D_%7B%28g%29%7D%2B14H_%7B2%7DO_%7B%28g%29%7D)
2. Find the limiting reagent:
- First calculate the number of moles of hexane and oxygen with the mass given by the problem.
For the hexane:
![70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}](https://tex.z-dn.net/?f=70.0gC_%7B6%7DH_%7B14%7D%2A%5Cfrac%7B1molC_%7B6%7DH_%7B14%7D%7D%7B86.2gC_%7B6%7DH_%7B14%7D%7D%3D0.81molesC_%7B6%7DH_%7B14%7D)
For the oxygen:
![81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}](https://tex.z-dn.net/?f=81.3gO_%7B2%7D%2A%5Cfrac%7B1molO_%7B2%7D%7D%7B32.0gO_%7B2%7D%7D%3D2.54molesO_%7B2%7D)
- Then divide the number of moles between the stoichiometric coefficient:
For the hexane:
![\frac{0.81}{2}=0.41](https://tex.z-dn.net/?f=%5Cfrac%7B0.81%7D%7B2%7D%3D0.41)
For the oxygen:
![\frac{2.54}{19}=0.13](https://tex.z-dn.net/?f=%5Cfrac%7B2.54%7D%7B19%7D%3D0.13)
- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.
3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:
The calculations must be done with the limiting reagent, that is the oxygen.
![81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}](https://tex.z-dn.net/?f=81.3gO_%7B2%7D%2A%5Cfrac%7B1molO_%7B2%7D%7D%7B32gO_%7B2%7D%7D%2A%5Cfrac%7B12molesCO_%7B2%7D%7D%7B19molesO_%7B2%7D%7D%2A%5Cfrac%7B44.0gCO_%7B2%7D%7D%7B1molCO_%7B2%7D%7D%3D70.6gCO_%7B2%7D)