Rozwiązanie 1:

Butan

$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$  

$2C_4{H}_{10}+9O_2\to 8\mathbf{C}\mathbf{O}+10H_{2}O$  

$2C_4{H}_{10}+5O_2\to 8\mathbf{C}+10H_{2}O$   

But-1-en

a)   $C_4{H}_8+6O_2\to 4C{O}_2+4H_{2}O$  

b)   $C_4{H}_8+4O_2\to 4\mathbf{C}\mathbf{O}+4H_{2}O$  

$C_4{H}_8+2O_2\to 4\mathbf{C}+4H_{2}O$  

But-1-yn

$2C_4{H}_6+11O_2\to 8C{O}_2+6H_{2}O$  

$2C_4{H}_6+7O_2\to 8\mathbf{C}{\mathbf{O}}_2+6H_{2}O$  

$2C_4{H}_6+3O_2\to 8\mathbf{C}+6H_{2}O$  

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Rozwiązanie 2:

Podczas spalania całkowitego powstaje tlenek węgla(IV) CO₂ i woda. Podczas spalania niecałkowitego powstaje tlenek węgla(II) CO i woda lub węgiel C i woda.

Butan

But-1-en

But-1-yn

 

Obliczenia:

Butan

$a)C_4{H}_{10}+\frac{13}{2}{O}_2\to 4C{O}_2+5H_{2}O\mid \cdot 2$  

 

$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$

$b)C_4{H}_{10}+\frac{9}{2}{O}_2\to 4CO+5H_{2}O\mid \cdot 2$  

$2C_4{H}_{10}+9O_2\to 8CO+10H_{2}O$  

$c)C_4{H}_{10}+\frac{5}{2}{O}_2\to 4C+5H_{2}O\mid \cdot 2$   

$2C_4{H}_{10}+5O_2\to 8C+10H_{2}O$   

But-1

  $a)C_4{H}_8+6O_2\to 4C{O}_2+4H_{2}O$  

$b)C_4{H}_8+4O_2\to 4CO+4H_{2}O$  

$c)C_4{H}_8+2O_2\to 4C+4H_{2}O$  

But-1-yn

$a)C_4{H}_6+\frac{11}{2}{O}_2\to 4C{O}_2+3H_{2}O\mid \cdot 2$  

$2C_4{H}_6+11O_2\to 8C{O}_2+6H_{2}O$  

$b)C_4{H}_6+\frac{7}{2}{O}_2\to 4CO+3H_{2}O\mid \cdot 2$   

$2C_4{H}_6+7O_2\to 8CO+6H_{2}O$  

$c)C_4{H}_6+\frac{3}{2}{O}_2\to 4C+3H_{2}O\mid \cdot 2$  

$2C_4{H}_6+3O_2\to 8C+6H_{2}O$