2. How would I calculate the equivalent weight of each of the following: a) KMnO

Question

2. How would I calculate the equivalent weight of each of the following:

a) KMnO4 reduced to MnO2

b) Ca(NO2)2 oxidized to nitrate

c) NaClO4 reduced to Cl-

3. How would I calculate the normality if 2.994 g of H2C2O4 * 2H2O is dissolved in 250.0 ml of solution?

4. How would I solve:

A 0.2412 gram sample of aluminum ore was dissolved in a dilute mineral acid. The solution was treated with an excess of sodium oxalate. The resultant aluminum oxalate precipitate was filtered, washed and redissolved in mineral acid, then titrated with 42.12 mL of 0.1098 N KMNO4. Calculate the percent aluminum in the sample.

Explanation / Answer

2.

a) KMn(VII)O4 reduced to Mn(IV)O2

Equivalent weight = M.wt/change in oxidation number. = M.wt/(7-4) = M.wt/3.

b) Ca(NO2)2 oxidized to nitrate NO2-(III) to NO3-(v)

=M.wt/(5-3) = M.wt/2.

c) NaClO4 reduced to Cl- ClO4-(+7) to Cl-(-1)

=M.wt/(7-(-1)) = M.wt/8.

3.

Normality N = wt/e.wt *1000/V in ml.

Equivalent weight = M.wt/Basicity of oxalicacid = 126/2 = 63

= 2.994/63*1000/250 = 0.19 N

4.

4Al + 6Na2(C2O4) ----> 2Al2(C2O4)3 + 6KMnO4 ----> 3K2(C2O4)
Aluminum oxalate and potassium permanganate are at a 1:3 ratio.

Moles of KMnO4
C = 0.1098 mol/L
v = 42.12 ml = 0.04212 L
C = n/v
n= CV = 0.004624776 mols
Therefore moles of aluminum oxylate = 0.001541592 mol = moles of aluminum
m= 26.98 g/mol * moles
mass of aluminum in oxylate = 0.04159215 g
% of initial sample = mass of aluminum / mass of ore
= 17.24% aluminum in ore sample

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