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Sagot :
Answer:
[tex]m_{KMnO_4}=90.9gKMnO_4[/tex]
Explanation:
Hello there!
In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:
[tex]m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2} *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4[/tex]
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