After the reaction how much octane is left

2 C8H18 + 25 O2 –> 16 CO2 + 18 H20 0.660 mol of octane is allowed to react with .780 mol of oxygen.
Oxygen is the limiting reactant.
After the reaction, how much octane is left?

General guidance

Concepts and reason

This is based on the concept that the limiting reagent decides how much amount of other reactants will be consumed when the reaction takes place. This decides the amount of product formed. Limiting reagent itself consumed fully and when it consumed, the reaction stops. This is only determined by the help of a balanced chemical equation. Limiting reagent also helps us to identify the percentage yield of the reaction.

Fundamentals

Limiting reagent is also called as limiting reactant. In a reaction, it is limiting reagent consumed totally and the remaining reactants are called as excess reagents.
Percentage theoretical yield is defined as the amount of the formation of the product when the limiting reagent is consumed fully.

Step-by-step

Step 1 of 3

Consider the reaction:

This reaction is balanced.

The given reaction is a balanced chemical equation since the number of moles of carbon, hydrogen and oxygen are same on both sides, i.e. 16 moles for carbon, 36 moles for hydrogen, and 50 moles for oxygen.

Make sure the chemical equation used in limiting reagent calculation should be balanced.

Also Read :   Chemistry: Which of the following is paramagnetic and which are diamagnetic?

Find out the number of moles of consumed.

Step 2 of 3

Calculation for the amount of is consumed is as follows:

Amount of octane () consumed is found out by dividing the number of the moles of octane involved (i.e. 2 moles) to the number of moles of limiting reagent involved (i.e. 25 moles) and multiply that number to the amount of limiting reagent consumed (i.e. 0.780 moles).

Make sure the number of moles of limiting reagent and excess reagent will only be taken from the balanced chemical reaction.

Step 3 of 3

Calculate the remaining moles as follows:

Moles of consumed is 0.598 moles.

Remaining moles of are calculated by subtracting the original moles of from the consumed moles of .

Answer

Moles of consumed is 0.598 moles.

Answer only

Moles of consumed is 0.598 moles.

2 C,H,g+250, ™16 CO, +18 H,0
0.780 moles of O, X 2 mol of CH,8 = 0,0624 mol of CH8 25 mol of 2
remaining = noriginal consumed = 0.660 mol-0.0624 mol =0.598 mol
CH, (Mcmaining)
CH: (noriginal)
C,H, (Mconsumed)
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After the reaction, how much octane is left?
2 C8H18 + 25 O2 --> 16 CO2 + 18 H20<br />
0.660 mol of octane is allowed to react with .780 mol of oxygen.<br />
Oxygen is the limiting reactant.<br />
After the reaction, how much octane is left?’/> <img src=

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