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Answer 1 · 2016-10-22T21:13:13

$"C"_10"H"_20"O"_2+14"O"_2 rarr 10"CO"_2 + 10"H"_2"O"$

The easiest way to do this is to balance the carbon and hydrogen atoms, and then balance the oxygen atoms.

$"C"_10"H"_20"O"_2+"O"_2 rarr "CO"_2 + "H"_2"O"$

We have $10$ $"C"$ atoms on the LHS, so we must have the same on the right hand side:

$"C"_10"H"_20"O"_2+"O"_2 rarr color(red)(10)"CO"_2 + "H"_2"O"$

We have $20$ $"H"$ atoms on the LHS, so that will give us $10$ $"H"_2"O"$ molecules:

$"C"_10"H"_20"O"_2+"O"_2 rarr color(red)(10)"CO"_2 + color(blue)(10)"H"_2"O"$

On the RHS we have $30$ $"O"$ atoms, and on the left, $2$ . This means we'll need $14$ $"O"_2$ atoms to balance the equation:

$"C"_10"H"_20"O"_2+color(green)(14)"O"_2 rarr color(red)(10)"CO"_2 + color(blue)(10)"H"_2"O"$