Solve Redox Titration Problem w/ NaNO2 & KMnO4 - Help Needed!

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SUMMARY

The discussion focuses on solving a redox titration problem involving potassium permanganate (KMnO4) and sodium nitrate (NaNO2). The balanced chemical equation for the reaction is MnO4- + BO2- + H+ → Mn2+ + NO3- + H2O. A 45 mL solution of 0.05 M KMnO4 completely reacts with 20.0 mL of NaNO2. To find the molar concentration of sodium nitrate, one must first calculate the number of moles of KMnO4 used, determine the stoichiometric amount of NaNO2, and then divide by the volume of the NaNO2 solution (0.02 L) to obtain the concentration.

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I have many problesm with this question. Can someone help me please?

In a titration of patasium permanganate soloution with acidified sodium nitrate solution

MnO4- + BO2- H+ -----> Mn+2 + NO3- +H2O

45 ml of 0.05M of KMnO4 reacts complitely with 20.0 mL of NaNO2 solution.

Calculate the molat concentration of sodium nitrate.

Please give me a hand with this. I relly need help to pass my chimetry curse.
 
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You must start with balanced equation. Once it is balanced calculate number of moles permanganate used and stoichiometric amount of sodium nitrate. Latter divided by 0.02L will give you concentration.
 
Sorry but I kinda dontt understand how to get the stoichiometric amount of sodium nitrate. And what do I have to divide by 0.02L. Sorry it's just I kinda not that good at chemestry.
 
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