Answer:
Option B. 0.0031 mole of C8H18
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Step 2:
Determination of the mass of C8H18 and the mass of O2 that reacted from the balanced equation.
This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g/mol
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 25 x 32 = 800g
From the balanced equation above, 228g of C8H18 reacted with 800g of O2.
Step 3:
Determination of the limiting reactant.
From the balanced equation above, 228g of C8H18 reacted with 800g of O2.
Therefore, 3g C8H18 will react with = (3 x 800) /228 = 10.53g of O2.
From the calculations made above, we can see that it will take a higher mass of 10.53g of O2 than what was given (i.e 9.25g) to react completely with 3g of C8H18.
Therefore, C8H18 is the excess reactant and O2 is the limiting reactant.
Step 4:
Determination of the excess mass of C8H18 that remains after the reaction.
This is illustrated below:
From the balanced equation above, 228g of C8H18 reacted with 800g of O2.
Therefore, Xg of C8H18 will react with 9.25g of O2 i.e
Xg of C8H18 = (228 x 9.25)/800
Xg of C8H18 = 2.64g
Therefore, 2.64g of C8H18 reacted.
The excess mass of C8H18 that remains can be obtained as follow :
Mass C8H18 = 3g
Mass of C8H18 that reacted = 2.64g
Mass of excess C8H18 that remains = (Mass C8H18) – (Mass of C8H18 that reacted)
Mass of excess C8H18 that remains = 3 – 2.64 = 0.36g
Step 5:
Conversion of the excess mass of C8H18 that rem to moles. This is illustrated below:
Mass of excess C8H18 that remains = 0.36g
Molar Mass of C8H18 = (12x8) + (18x1) = 114g/mol
Number of mole = Mass/Molar Mass
Number of mole of C8H18 that remains = 0.36/114 = 0.0031 mole
Therefore, 0.0031 mole of C8H18 remained
