Answer:
[tex]C_1_0H_7OH (l) + 23/2 O_2 (g) \longrightarrow 10 CO_2 (g) + 4 H_2O (g)[/tex]
Explanation:
Hi, the reaction you asked for is the complete combustion of 1-naphthol in which the naphthol reacts with oxygen producing water and carbon dioxide:
[tex]C_1_0H_7OH (l) + O_2 (g) \longrightarrow CO_2 (g) + H_2O (g)[/tex]
Steps to balance:
1) Number of [tex]CO_2[/tex] equal to the C in the naphthol:
[tex]C_1_0H_7OH (l) + O_2 (g) \longrightarrow 10 CO_2 (g) + H_2O (g)[/tex]
2) Number of [tex]H_2O[/tex] equal to half the H in the naphthol:
[tex]C_1_0H_7OH (l) + O_2 (g) \longrightarrow 10 CO_2 (g) + 4 H_2O (g)[/tex]
2) Balance the [tex]O_2[/tex] equal to O in [tex]CO_2[/tex] and [tex]H_2O[/tex] minus the O in the naphthol :
[tex]C_1_0H_7OH (l) + 23/2 O_2 (g) \longrightarrow 10 CO_2 (g) + 4 H_2O (g)[/tex]
If you prefer, write the equation in non-fractional numbers:
[tex]2 C_1_0H_7OH (l) + 23 O_2 (g) \longrightarrow 20 CO_2 (g) + 8 H_2O (g)[/tex]
The balanced chemical equation for the combustion of 1-naphthol, C₁₀H₇OH, to give CO₂ and H₂O is:
A combustion reaction is a reaction in which a substance is burned in air i.e oxygen to produce carbon (iv) oxide and water.
The equation for the combustion of 1-naphthol, C₁₀H₇OH can be written as follow:
C₁₀H₇OH + O₂ —> CO₂ + H₂O
The equation above can be balanced as follow:
C₁₀H₇OH + O₂ —> CO₂ + H₂O
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:
C₁₀H₇OH + O₂ —> 10CO₂ + H₂O
There are 8 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 4 before H₂O as shown below:
C₁₀H₇OH + O₂ —> 10CO₂ + 4H₂O
There are a total of 24 atoms of O on the right side and 3 atoms on the left side. It can be balance by writing [tex]\frac{23}{2}[/tex] before O₂ as shown below:
C₁₀H₇OH + [tex]\frac{23}{2}[/tex]O₂ —> 10CO₂ + 4H₂O
Multiply through by 2 to clear the fraction
2C₁₀H₇OH + 23O₂ —> 20CO₂ + 8H₂O
Thus, the equation is balanced.
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