How do you balance $C_{6} H_{14} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$ $C_6H_14(g) + O_2(g) -> CO_2(g) + H_2O(l)$ ?

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How do you balance $C_{6} H_{14} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$ $C_6H_14(g) + O_2(g) -> CO_2(g) + H_2O(l)$ ?

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Answer 1 · 2016-02-20T09:41:02

Garbage in MUST equal GARBAGE out.

Explanation:

$C_{6} H_{14} (l) + \frac{19}{2} O_{2} (g) \to 6 C O_{2} (g) + 7 H_{2} O (l)$ $C_6H_14(l) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(l)$

Is this balanced with respect to mass? Look at the left hand side. There are 6 carbon atoms, 14 hydrogens atoms, and 19 oxygen atoms. On the right hand side, there are 6 carbon atoms........ If there is not equality of mass on RHS and LHS, you know for a fact it is unbalanced.

How does energy transfer in this reaction?

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How do you balance $C_{6} H_{14} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$ $C_6H_14(g) + O_2(g) -> CO_2(g) + H_2O(l)$ ?

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How do you balance C6H14(g)+O2(g)→CO2(g)+H2O(l)C_6H_14(g) + O_2(g) -> CO_2(g) + H_2O(l)?

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How do you balance $C_{6} H_{14} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$ $C_6H_14(g) + O_2(g) -> CO_2(g) + H_2O(l)$ ?