How do you balance #C_10H_16 + Cl_2 -> C + HCl#?

1 Answer
Jun 2, 2016

Answer:

#C_10H_16 + 8 Cl_2 →10 C + 16 HCl#

Explanation:

With problems like this, I would do an atom inventory to see how many atoms of each element are present on both sides of the reaction arrow.

Initially, you have 10 atoms of C, 16 atoms of H, and 2 atoms of Cl on the reactants side, followed by 1 atom of C, 1 atom of H, and 1 atom of Cl on the products side.

Balance the carbon first by putting a coefficient of 10 in front of the C on the products side. Afterwards, balance the H by putting a coefficient of 16 in front of the HCl molecule. Now, the only thing that's not balanced is chlorine. So just place a coefficient of 8 in front of the #Cl_2# molecule, since you obtain 16 when the subscript is multiplied by the coefficient.

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