If 220 grams of C3H8 react with 20 grams of O2, the amount of CO2 formed in the process is 16.5 grams of CO2
The equation for this reaction will be as follows:
[tex]\mathbf{C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O}[/tex]
From the given information:
The number of moles of C₃H₈ present can be computed as:
[tex]\mathbf{no \ of \ mole s= \dfrac{mass}{molar \ mass}}[/tex]
[tex]\mathbf{no \ of \ mole s= \dfrac{220 \ g}{44 \ g/mol}}[/tex]
[tex]\mathbf{no \ of \ moles \ of \ C_3H_8=5 \ moles}[/tex]
The number of moles of O₂ present in the reaction is:
[tex]\mathbf{no \ of \ mole s= \dfrac{20 \ g}{32 \ g/mol}}[/tex]
[tex]\mathbf{no \ of \ moles\ O_2= 0.625 \ moles}[/tex]
Now, from the above-balanced chemical equation, the amount of CO₂ that can be formed depends largely on the amount of O₂ present.
As such;
- the number of moles of CO₂ formed can be estimated as:
[tex]\mathbf{= \dfrac{0.625 \ moles \ of \ O_2 \times 3 \ moles \ of \ CO_2 }{5 \ moles \ of \ O_2}}[/tex]
[tex]\mathbf{= 0.375 \ moles \ of \ CO_2 \ is \ formed}[/tex]
Since molar mass of CO2 = 44.01 g/mol
The mass of CO2 in grams = 0.375 moles × 44.01 g/mol
= 16.5 grams of CO2
Therefore, we can conclude that the amount of CO2 formed in the process is 16.5 grams of CO2.
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