Respuesta :
There were 0.384 g [tex]H_{2} O_{2}[/tex] in the sample.
[tex]\frac{43.2 ml of MnO_{4} ^{-} }{1}[/tex] X[tex]\frac{0.105 m dos MnO^{-} _{4} }{1000 ml MnO^{-}_{4} }[/tex] x [tex]\frac{5 mol H_{2} O_{2} }{2 mol MnO^{-}_{4} }[/tex] = 0.0113 mol [tex]H_{2} O_{2}[/tex]
[tex]\frac{0.0113 mol H_{2}O_{2} }{1}[/tex] X [tex]\frac{34.02g}{1 mol}[/tex] = 0.384 g of
A chemical process known as oxidation-reduction is taking on here, which includes the exchange of electrons between two distinct species. Thus, the change in our species' oxidation states was produced by the transfer of electrons. Since we are concentrating on hydrogen peroxide in this instance, we have 14.8 g of [tex]H_{2} O_{2}[/tex] that was measured with a close ratio [tex]MnO^{-} _{4}[/tex]. We have 0.113 moles of [tex]H_{2} O_{2}[/tex]. The mass percent of [tex]H_{2} O_{2}[/tex] in the sample is 2.6 and the grams of [tex]H_{2} O_{2}[/tex] in the sample are 0.384 g.
Learn more about oxidation-reduction reaction: https://brainly.com/question/19528268
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