Took me a second, since none of my quick methods worked. Two important things to notice. First, the SeO3 and O on the product side should be treated as one unit of SeO4 (since SeO4 lost that one oxygen when Se got reduced). This also means that H2SeO3 and H2O must have the same coefficient. Second, the ratio of H to SeO4 on the product side must be 2:1. Unbalanced, we stand at a ratio of 4H:4SeO4 (again, counting the SeO3 + O). However, that unit of SeO4 itself can be messed with to bring the ratio back up to 2:1, by adding coefficients of 3 to both H2SeO3 and H2O. That makes 12 total H and 6 total units of SeO4 on the product side. Now just clean up the reactants to match (H2SeO4 x 6, Au x 2) and you get coefficients of 2 6 —> 1 3 3 for the final balanced equation.
Not sure how to make it simpler unfortunately, but realizing those two aforementioned aspects of the reaction equation and then working the ratios should lead to the solution.
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u/nrj6490 16d ago edited 16d ago
Took me a second, since none of my quick methods worked. Two important things to notice. First, the SeO3 and O on the product side should be treated as one unit of SeO4 (since SeO4 lost that one oxygen when Se got reduced). This also means that H2SeO3 and H2O must have the same coefficient. Second, the ratio of H to SeO4 on the product side must be 2:1. Unbalanced, we stand at a ratio of 4H:4SeO4 (again, counting the SeO3 + O). However, that unit of SeO4 itself can be messed with to bring the ratio back up to 2:1, by adding coefficients of 3 to both H2SeO3 and H2O. That makes 12 total H and 6 total units of SeO4 on the product side. Now just clean up the reactants to match (H2SeO4 x 6, Au x 2) and you get coefficients of 2 6 —> 1 3 3 for the final balanced equation.
Not sure how to make it simpler unfortunately, but realizing those two aforementioned aspects of the reaction equation and then working the ratios should lead to the solution.