9.525 moles of O₂
Given:
Combustion of 88.9 g of C₂H₄ to form CO₂ and H₂O.
Question:
How many moles of O₂ are required for the complete reaction of combustion of C₂H₄?
The Process:
Let us convert mass to mole for C₂H₄.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{88.9}{28} = 3.175 \ moles \ }[/tex]
The combustion reaction of C₂H₄ (ethylene, also named ethene) can be expressed as follows:
[tex]\boxed{ \ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \ }[/tex] (the reaction is balanced)
According to chemical equation above, proportion between C₂H₄ and O₂ is 1 to 3. Therefore, we can count the number of moles of O₂.
[tex]\boxed{ \ \frac{n(O_2)}{n(C_2H_4)} = \frac{3}{1} \ }[/tex]
[tex]\boxed{ \ n(O_2) = \frac{3}{1} \times n(C_2H_4) \ }[/tex]
[tex]\boxed{ \ n(O_2) = \frac{3}{1} \times 3.175 \ moles \ }[/tex]
Thus, the number of moles of O are required for the complete reaction of the combustion of C₂H₄ is 9.525 moles.
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Notes:
If we want to calculate the mass of O₂, then we use the number of moles of O₂ that have been obtained.
