When 10 moles of O₂ reacts with 1.4 moles of C₁₀H₈ what is the limiting reactant? *

O2 is the limiting reactant.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C10H18 + 29O2 —> 20CO2 + 18H2O

The limiting reactant can be obtained as follow:

From the balanced equation above,

2 moles of C10H18 reacted with 29 moles of O2.

Therefore, 1.4 mole of C10H18 will react with = (1.4 x 29)/2 = 20.3 moles of O2.

From the above calculation, we can see that it will take a higher amount of O2 i.e 20.3 moles than what was given i.e 10 moles to react completely with 1.4 mole of C10H18. Therefore, O2 is the limiting reactant and C10H18 is the excess reactant.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

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