Answer:
[tex]2C_{10} H_{22(l)} + 31O2(g)--> 20CO_{2(g)} + 22H_{2}O_{(g)}[/tex]
Explanation:
[tex]2C_{10} H_{22(g)} + 31O2(g)--> 20CO_{2(g)} + 22H_{2}O_{(g)}[/tex]
From the question, one can work out which states of matter to assign to which species. The trick with organic equations of this nature is to try to balance everything but oxygen first. Make sure you balance oxygen last because it is the easiest to balance.
The balanced chemical equation for the reaction between liquid decane and oxygen gas to form carbon dioxide gas and water vapor is:
The chemical equation for the reaction between liquid decane and oxygen gas to form carbon dioxide gas and water vapor can be written as follow:
Decane => C₁₀H₂₂
Oxygen gas => O₂
Carbon dioxide => CO₂
Water => H₂O
Decane + Oxygen gas —> Carbon dioxide + Water
C₁₀H₂₂(l) + O₂(g) —> CO₂(g) + H₂O(g)
The above equation can be balance as follow:
There are 22 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 11 before H₂O as shown below:
C₁₀H₂₂(l) + O₂(g) —> CO₂(g) + 11H₂O(g)
There are 10 atoms of C on the left side and 1 atom on the right side. It can be balance by writing 10 before CO₂ as shown below:
C₁₀H₂₂(l) + O₂(g) —> 10CO₂(g) + 11H₂O(g)
There are a total of 31 atoms of O on the right side and 2 atoms on the left side. It can be balance by writing [tex]\frac{31}{2}[/tex] before O₂ as shown below:
C₁₀H₂₂(l) + [tex]\frac{31}{2}[/tex]O₂(g) —> 10CO₂(g) + 11H₂O(g)
Multiply through by 2 to express in whole number.
2C₁₀H₂₂(l) + 31O₂(g) —> 20CO₂(g) + 22H₂O(g)
Now, the equation is balanced.
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