How do you balance $C_{4} H_{8} + O_{2} \to C O_{2} + H_{2} O$ $C_4H_8 + O_2 -> CO_2 + H_2O$ ?

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Answer 1 · 2016-12-22T03:48:52

$Garbage in equals garbage out.......$ $"Garbage in equals garbage out......."$

The usual rigmarole is (i) balance the carbons, (ii) then the hydrogens, and (iii) then the oxygens.

$C_{4} H_{8} + 6 O_{2} \to 4 C O_{2} (g) + 4 H_{2} O (g)$ $C_4H_8 + 6O_2 rarr 4CO_2(g) + 4H_2O(g)$

Can you do same for $pentane$ $"pentane"$ , $C_{5} H_{12}$ $C_5H_12$ ? It is a little harder to balance.

How do you balance C_4H_8 + O_2 -> CO_2 + H_2OC4H8+O2→CO2+H2OC_4H_8 + O_2 -> CO_2 + H_2O?