Answer :
Taking into account the reaction stoichiometry, 6.15 grams of COâ‚‚ are formed when 8.94 grams of Oâ‚‚ reacts completely.
In first place, the balanced reaction is:
CH₄ + 2 O₂ → CO₂ + 2 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:
- CHâ‚„: 1 mole
- Oâ‚‚: 2 moles
- COâ‚‚: 1 mole
- Hâ‚‚O: 2 moles
The molar mass of the compounds is:
- CHâ‚„: 16.04 g/mole
- Oâ‚‚: 32 g/mole
- COâ‚‚: 44.01 g/mole
- Hâ‚‚O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CH₄: 1 mole× 16.04 g/mole= 16.04 grams
- O₂: 2 moles× 32 g/mole= 64 grams
- CO₂: 1 mole× 44.01 g/mole= 44.01 grams
- H₂O: 2 moles× 18 g/mole= 36 grams
Then the following rule of three can be applied: if by reaction stoichiometry 64 grams of Oâ‚‚ form 44.01 grams of COâ‚‚, 8.94 grams of Oâ‚‚ form how much mass of COâ‚‚?
[tex]mass of CO_{2} =\frac{8.94 grams of O_{2} x44.01 grams of CO_{2} }{64 grams of O_{2}}[/tex]
mass of COâ‚‚= 6.15 grams
Finally, 6.15 grams of COâ‚‚ are formed when 8.94 grams of Oâ‚‚ reacts completely.
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