in the reaction a^n+ + mno4- gives a^+5 + mn+2 if 0.05 mole of a^+n is oxidized by 0.02 moles of mno4- then the value of n is
Question
in the reaction a^n+ + mno4- gives a^+5 + mn+2 if 0.05 mole of a^+n is oxidized by 0.02 moles of mno4- then the value of n is
Solution
To find the value of n, we need to use the stoichiometry of the reaction.
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First, we need to balance the reaction. The balanced reaction is:
a^n+ + 5MnO4^- -> a^+5 + 5Mn^+2 + 4O2
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From the balanced equation, we can see that one mole of a^n+ reacts with five moles of MnO4^- to produce one mole of a^+5 and five moles of Mn^+2.
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Given that 0.05 moles of a^n+ is oxidized by 0.02 moles of MnO4^-, we can set up the following ratio:
0.05 moles a^n+ / 0.02 moles MnO4^- = 1 mole a^n+ / 5 moles MnO4^-
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Solving for n, we get:
n = 5 * (0.05 / 0.02) = 12.5
Therefore, the value of n is 12.5.
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