Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O.
Suppose 4.6 g of octane is mixed with 27.4 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

14.20g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O

Step 2:

Determination of the masses of CH3(CH2)6CH3 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below

Molar Mass of CH3(CH2)6CH3 = 12 + (3x1) + 6[12 + (2x1)] + 12 + (3x1) = 12 + 3 + 6[12 + 2] + 12 + 3 = 12 + 3 + 6[14] + 12 + 3 = 114g/mol

Mass of CH3(CH2)6CH3 from the balanced equation = 2 x 114 = 228g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

From the balanced equation above, 228g of CH3(CH2)6CH3 reacted with 800g of O2 to produce 704g of CO2.

Step 3:

Determination of the limiting reactant. We must obtain the limiting reactant in order to get the maximum yield of CO2.

The limiting reactant is obtained as follow:

From the balanced equation above, 228g of CH3(CH2)6CH3 reacted with 800g of O2.

Therefore, 4.6g will react with = (4.6x800)/228 = 16.14g of O2.

The mass of O2 that reacted is lesser than the given mass ( i.e 27.4g) from the question. Therefore, CH3(CH2)6CH3 is the limiting reactant.

Step 4:

Determination of the maximum mass of the CO2 produced.

The limiting reactant is used to obtain the maximum yield in any reaction.

From the balanced equation above, 228g of CH3(CH2)6CH3 produce 704g of CO2.

Therefore, 4.6g of CH3(CH2)6CH3 will produce = (4.6 x 704)/228= 14.20g of CO2.

Therefore, the maximum mass of CO2 produced is 14.20g

answer

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