Determine the well balanced chemical equation because that this reaction. C8H18(g)+O2(g)→CO2(g)+H2O(g)

Part A:Enter the coefficients because that each compound in order, separated by commas. Because that example, 1,2,3,4 would suggest one mole that C8H18, 2 moles the O2, three moles that CO2, and also four mole of H2O.

You are watching: C8h18 + o2 = co2 + h2o


Part B:0.280 mol of octane is permitted to react v 0.630 mol that oxygen. I m sorry is the limiting reactant?Oxygen

Part C:How numerous moles the water are produced in this reaction?


Part D:After the reaction, exactly how much octane is left?

Concepts and also reasonBalance the equation by placing proper coefficients prior to the reactant and products. Then compose the coefficients the present prior to the each compound.Use mole ratio ide to discover the limiting reagent. Moles of one substance to mole of an additional substance in a well balanced chemical equation is dubbed mole ratio.Calculate the number of moles of water by utilizing the moles of the limiting reagent.

FundamentalsIn a balanced chemical equation, number of atoms of each element and total in the reactant side and product side are equal to each other.A limiting reagent is totally consumed in a chemistry reaction.A conversion element is used to convert between moles the one problem to mole of an additional substance.


Part A:The well balanced equation is as follows:

The coefficients the the each link in order room 2, 25, 16, and 18.



Part B:Calculate number of moles that

by making use of the moles of every reactant.
The reactant oxygen create less product contrasted to the reactant octane. So, the limiting reagent is .

Part BLimiting reagent is .

Compare the moles calculated by making use of the reactants oxygen and also octane. Limiting reagent is one that has actually less number of moles compared with the other reactants in the reaction. The reactant . Less variety of moles, for this reason it will certainly be the limiting reagent.

Part C:


The reactant because that which we calculate moles have to be ~ above the peak of the counter factor.Moles of must be top top the top of the counter factor. The calculated moles of is 0.453 mol.

Part D:

Calculate mole of octane reacted by using the following conversion factor.Conversion variable to gain moles that octane from moles of octane is together follows:


Part DAmount of octane left is 0.229mol.

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The reactant because that which we calculate moles need to be top top the height of the conversion factor.Moles the octane should be on the top of the conversion factor. Calculate the lot of octane continued to be by subtracting the mole of octane spend from the mole of octane allowed in the reaction.The calculated mole the octane left in the reaction is 0.229mol.