Mass Average NaNO2 mo Molarity o.oviyON Number of electrons gained b electrons i

Question

Mass Average NaNO2 mo Molarity o.oviyON Number of electrons gained b electrons in this reaction KMno by one formula Mr Normality Show normality KMnoa 60202 Calculation 1st Try 2nd Try 3rd Try Average FINAL reading of Buret, mL (after end point is reached) I4 INITAL reading of Buret, mL. VOLUME KMno4 mL, used to reach end point it t Experimental value for Gram Theoretical value of Gram error Equivalent Weight of NaNO2 Equivalent Weight of NaNO2 Show calculation: Show calculation: Show calculation: asso mote in OY weight Exp. 17.5

Explanation / Answer

Chemical equation,

5NaNO2 + 2KMnO4 + 3H2SO4 ---> 5NaNO3 + 2MnSO4 + K2SO4 + 3H2O

For (1),

mass NaNO2 = 0.164 g

molarity of KMnO4 solution = 0.0446 M

Number of electrons involved = 5

Normality of KMnO4 = 0.0446 M x 5 = 0.223 N

Average volume of KMnO2 dispensed = 19 ml

moles of KMnO4 used = molaity x volume = 0.0446 M x 0.019 L = 0.0008474 mol

moles of NaNO2 reacted = 0.0008474 mol x 5/2 = 0.00212 mol

NaNO2 --> NaNO3

number of electron involved = 2

Experimental gram NaNO2 reacted = 0.00212 mol x 69 g/mol = 0.1463 g

Expriemental gram equivalent weight = 0.1463/2 = 0.07315

theoretical gram equivalent weight for NaNO2 = 0.164/2 = 0.082

Similarly other values of Mass of NaNO2 may be used to calculate the data.

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