Jawaban:

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:

\mathbf{C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O}C

3

H

8

+5O

2

→3CO

2

+4H

2

O

From the given information:

The number of moles of C₃H₈ present can be computed as:

\mathbf{no \ of \ mole s= \dfrac{mass}{molar \ mass}}no of moles=

molar mass

mass

\mathbf{no \ of \ mole s= \dfrac{220 \ g}{44 \ g/mol}}no of moles=

44 g/mol

220 g

\mathbf{no \ of \ moles \ of \ C_3H_8=5 \ moles}no of moles of C

3

H

8

=5 moles

The number of moles of O₂ present in the reaction is:

\mathbf{no \ of \ mole s= \dfrac{20 \ g}{32 \ g/mol}}no of moles=

32 g/mol

20 g

\mathbf{no \ of \ moles\ O_2= 0.625 \ moles}no of moles O

2

=0.625 moles

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:

\mathbf{= \dfrac{0.625 \ moles \ of \ O_2 \times 3 \ moles \ of \ CO_2 }{5 \ moles \ of \ O_2}}=

5 moles of O

2

0.625 moles of O

2

×3 moles of CO

2

\mathbf{= 0.375 \ moles \ of \ CO_2 \ is \ formed}=0.375 moles of CO

2

is formed

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.