Liquid octane (CH,(CH,) CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (Co,) and gaseous water (H,0). Suppose 6.9 g of octane is mixed with 39.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Liquid octane (CH,(CH,) CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (Co,) and gaseous water (H,0). Suppose 6.9 g of octane is mixed with 39.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Chemical Reaction Involving Octane and Oxygen**
Liquid octane \((CH_3 (CH_2)_6 CH_3)\) will react with gaseous oxygen \((O_2)\) to produce gaseous carbon dioxide \((CO_2)\) and gaseous water \((H_2O)\). Suppose 6.9 g of octane is mixed with 39.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
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This problem involves the combustion of octane, a hydrocarbon, with oxygen, a typical exothermic reaction releasing energy. The objective is to determine the mass of carbon dioxide produced under the given conditions.
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Expert Solution
Step 1
The combustion reaction of octane i.e CH3(CH2)6CH3 can be written as,
=> CH3(CH2)6CH3 + O2 --------> CO2 + H2O
Balancing : Since we have 8 C in LHS. Hence making it 8 in RHS also.
=> CH3(CH2)6CH3 + O2 --------> 8 CO2 + H2O
Now we have 18 H in LHS. Hence making it 18 in RHS also.
=> CH3(CH2)6CH3 + O2 --------> 8 CO2 + 9 H2O
Since total number of O in RHS is 25 now. Hence making it 25 in LHS also.
=> CH3(CH2)6CH3 + 12.5 O2 --------> 8 CO2 + 9 H2O
Multiplying the whole reaction with 2 to get lowest integral coefficient as.
=> 2 CH3(CH2)6CH3 + 25 O2 --------> 16 CO2 + 18 H2O
Since all the elements are in equal number in both side of the reaction now. Hence the above reaction is now balanced.
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