Respuesta :
You need to know the number of molecules of each reactant at start.
I will do an example for you, assuming you start with 10 molecules of O2 and 3 molecules of C2H5OH.
1) State the balanced chemical equation
C2H5OH + 3O2 ---> 2CO2 + 3H2O
=> molar ratios: 1 mol C2H5OH : 3 mol O2 : 2 mol CO2 : 3 mol H2O
2) Find the limiting reactant:
3 C2H5OH / 10 O2 < 1 C2H5OH / 3 O2
=> the 3 molecules of C2H5OH will react with 9 molecules of O2 ; 1 molecule of O2 will remain.
2) State the molar conversion
use the molar ratios to calculate the number of molecules of CO2 and O2 that will be produced.
This table shows the results:
CH2H5OH O2 CO2 H2O
start 3 10 0 0
react 3 9 - -
produced - - 6 9
--------------------------------------------------------
end 3-3 = 0 10-9 = 1 6 9
So, starting with 3 molecules of C2H5OH and 10 molecules of O2, when the reaction goes to completion, there will be present 0 molecules of C2H5OH, 1 molecule of O2, 6 molecules of CO2 and 9 molecule of H2O.
I will do an example for you, assuming you start with 10 molecules of O2 and 3 molecules of C2H5OH.
1) State the balanced chemical equation
C2H5OH + 3O2 ---> 2CO2 + 3H2O
=> molar ratios: 1 mol C2H5OH : 3 mol O2 : 2 mol CO2 : 3 mol H2O
2) Find the limiting reactant:
3 C2H5OH / 10 O2 < 1 C2H5OH / 3 O2
=> the 3 molecules of C2H5OH will react with 9 molecules of O2 ; 1 molecule of O2 will remain.
2) State the molar conversion
use the molar ratios to calculate the number of molecules of CO2 and O2 that will be produced.
This table shows the results:
CH2H5OH O2 CO2 H2O
start 3 10 0 0
react 3 9 - -
produced - - 6 9
--------------------------------------------------------
end 3-3 = 0 10-9 = 1 6 9
So, starting with 3 molecules of C2H5OH and 10 molecules of O2, when the reaction goes to completion, there will be present 0 molecules of C2H5OH, 1 molecule of O2, 6 molecules of CO2 and 9 molecule of H2O.
Molecules of CO₂, H₂O, C₂H₅OH, and O₂ will be present if the reaction goes to completion are
0 molecules C₂H₅OH
0 O₂ molecules
2 and CO₂ molecules
3 H₂O molecules
Further explanation
The reaction equation is the chemical formula of reagents and product substances
the reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
The reaction of the combustion of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Equivalent reaction equation, if the number of atomic substances of the reactor is the same as the product substance
The reaction coefficient also shows the ratio of moles, volumes or molecules of substances involved in the reaction
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion resulting in CO and H₂O
In perfect combustion of ethanol (C₂H₅OH) and oxygen (O₂), ethanol and oxygen will be converted entirely into CO₂ and H₂O at the end of the reaction
Ethanol and oxygen reactions:
C₂H₅OH + O₂ ---> CO₂ + H₂O
We equalize first by giving coefficients a, b and c
C₂H₅OH + aO₂ ---> bCO₂ + cH₂O
C = left 2, right b -> b = 2
H = left 6, right 2c ---> c = 3
O = left 2a + 1, right 2b + c
2a + 1 = 2 (2) + 3
2a + 1 = 4 + 3
2a = 6
a = 3
So the equivalent reaction equation =
C₂H₅OH + 3O₂ ---> 2CO₂ + 3H₂O
The comparison of the coefficients shows that: 1 C₂H₅OH molecule reacts with 3 molecule O₂ produces 2 molecules of CO₂ and 3 molecules of H₂O
So the complete combustion of ethanol at the end of the reaction composition:
0 molecules C₂H₅OH
0 O₂ molecules
2 and CO₂ molecules
3 H₂O molecules
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Keywords : combustion, ethanol, molecules