If 28.70 Ml Of 0.0200 M

Calculate the percentage of C2O4 2 in each of the following:? 3

a) H2C2O4

B) Na2C2O4

c) K2C2O4

d) K3 [Al (C2O4) 3] x 3H2O

If 28.70 ml of 0.0200 M KMnO4 is required to estimate 0.250 g of K3 [Fe (C2O4) 3] x H2O sample, what is the percentage of C2O4

Divide the formula of oxalate m (88.01 g / mol) by molecular formula or mixture m and multiply by 100.

Applies to the following one:

(88.01 g / mol) / (90.03 g / mol) * 100 = 97.76%

Use MM or FM appropriately for each other.

The equilibrium equation for permangate / oxalate is:

2 MnO4 (water) + 5 C2O42 (water) + 16 H + (water)> 2 Mn2 + (water) + 10 CO2 (g) + 8 H2O (l)

Two permanganates react with 5 oxalate.

0.0200 M is 28.70 ml of KMnO4.

28.70 ml * 0.0200 mol / L * 0.001 L / ml = 0.000574 KMnO4

This means that 0.000574 * 5/2 = 0.001435 oxalate will be estimated.

88.01 grams / mol * 0.001435 = 0.1263 grams are oxalate.

This is 0.1263 grams / 0.250 grams complex * 100 = 50.52%.

HN3 H + + N3 Co__0__0_ CoX_____X___X_ Ka = X * X / (■■■)? X 2 / Co then X = (Co * one million 9x10 5) one million / 2 pc Ionization = (X / Co) * one Dred = = (one million 9x10 5 / Co) one Dredger Million / 2 *

If 28.70 Ml Of 0.0200 M