![]() ![]() Now, we have two options to even out the right hand side: We can either multiply C 6H 12O 6 or O 2 by a coefficient. On the right, there are eight oxygen molecules. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). So, we will add a coefficient of six on the hydrogen-containing molecule on the left. ![]() There are two hydrogen atoms on the left and twelve on the right. So, we add a coefficient of six on the carbon-containing molecule on the left. There is only one atom of carbon on the left hand side, but six on the right hand side. Here, both carbon and hydrogen fit this requirement. The first step is to focus on elements that only appear once on each side of the equation.
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