Answer :
The percent yield of O[tex]_{2}[/tex] in the given reaction is 49.6%.
To calculate the percent yield, we need to compare the actual yield of O[tex]_{2}[/tex] (38.6g) to the theoretical yield of O[tex]_{2}[/tex]. From the balanced equation, we can see that for every 2 moles of H[tex]_{2}[/tex]O[tex]_{2}[/tex] decomposed, 1 mole of O[tex]_{2}[/tex] is produced. Therefore, the theoretical yield of O2 can be calculated using the given amount of H[tex]_{2}[/tex]O[tex]_{2}[/tex] (3 mol).
The molar mass of O[tex]_{2}[/tex] is 32.00 g/mol, so the theoretical yield of O[tex]_{2}[/tex] is (3 mol H[tex]_{2}[/tex]O[tex]_{2}[/tex]) * (1 mol O[tex]_{2}[/tex]/2 mol H[tex]_{2}[/tex]O[tex]_{2}[/tex]) * (32.00 g O[tex]_{2}[/tex]/mol) = 48.00 g O[tex]_{2}[/tex].
Now we can calculate the percent yield using the formula: (actual yield/theoretical yield) * 100. Plugging in the values, we get (38.6g O[tex]_{2}[/tex]/48.00g O[tex]_{2}[/tex]) * 100 = 80.42%. Rounding to the nearest tenth, the percent yield of O[tex]_{2}[/tex] is 49.6%.
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