Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Tom on July 01, 2004, 04:35:30 PM
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The basic solution, MnO4 ¯ oxidizes NO2¯ to NO3¯ and is reduced to MnO2. What volume of 0.10 M KmnO4 would be needed to oxidize 30ml of a 0.10 M NaNO2 solution?
My conclusion came out that it would require 40ml of solution, does that sound logical? Again the all the numbers are subscripted other than the 0.10M and 30ml.
Thanks
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I suggest starting by writing the balanced chemical equation.
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I used this formula
D=M divided by V
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I don't think that'll do it. You'll have to write the balanced chemical formula.