Respuesta :
Answer:
6.3 grams of C2H6O
Explanation:
Using stoichiometry of the balanced equation, 12.0g CO2 *[tex]\frac{1mol CO_2}{44 grams CO_2}*\frac{1 mol C_2H_6O}{2 mol CO_2}*\frac{46 grams C_2H_6O}{1 mole C_2H_6O}[/tex]=6.3 grams of C2H6O
Considering the reaction stoichiometry, the mass of C₂H₆O necessary to produce 12.0 g CO₂ is 6.27 grams.
The balanced reaction is:
C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- C₂H₆O: 1 mole
- O₂: 3 moles
- CO₂: 2 moles
- H₂O: 3 moles
The molar mass of the compounds present in the reaction is:
- C₂H₆O: 46 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:
- C₂H₆O: 1 mole× 46 g/mole= 46 grams
- O₂: 3 moles× 32 g/mole= 96 grams
- CO₂: 2 moles× 44 g/mole= 88 grams
- H₂O: 3 moles× 18 g/mole= 54 grams
Then, it is possible to determine the mass of C₂H₆O necessary by the following rule of three: If by stoichiometry 88 grams of CO₂ are produced from 46 grams of C₂H₆O, 12 grams of CO₂ are produced from how much mass of C₂H₆O?
[tex]mass of C_{2} H_{6} O=\frac{12 grams of CO_{2}x 46 grams of C_{2} H_{6} O}{88 grams of CO_{2}}[/tex]
mass of C₂H₆O= 6.27 grams
Finally, the mass of C₂H₆O necessary to produce 12.0 g CO₂ is 6.27 grams.
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