Question

$use the one on the right for run2 solution$

Answer 1

Using the data above and the balanced reaction: Mno. 5 F2.8 Mn-2.5 Fe + 4H0. Clearly calculate the concentration of the KMnO. (MnO:) for both runs and calculate the average concentration of Mno. Make sure to include units.

=392,14 gr/mol

Molar Mass of Fe(NH)2(506)26H40 (read off the bottle.)

Run 1

Mass 1=0.301 g

WM=392,14g/mol

1 mol 19 * 392.14 g = 0,000768molle

Now with the equation

MnO- +5Fe2+ + 8H+ → Mn+2 + 5Fe3+ + 4H20

1 mol Mn 0,000768molFe* = 0,0001535molMn 5mol Fe

AV = V2 - V1 = 10.51 ml -0 ml = 10.51ml = 0.0105 L

MolMn04 Molarity KMnO4 = AV

0.0001535 mol Molarity KMnO4 = - 0.0195L

Molarity KMnO4 = 0.0146M

Run 2

Mass 1=0.324 g

WM=392,14g/mol

1 mol 0.324g * 392 140 = 0,000826molFe

Now with the equation

MnO- +5Fe2+ + 8H+ → Mn+2 + 5Fe3+ + 4H20

1 mol Mn 0,000826molFe* P = 0,0001652molan 5mol Fe

AV = V2 - V1 = 10.75 ml – 0.05 ml = 10.70ml = 0.0107 L

MolMn04 Molarity KMnO4 = AV

0.0001652 mol Molarity KMnO4 =- 0.0107L

Molarity KMnO4 = 0.0154M

To calculate the average concentration we add both molarities and divide by two

M1 + M2 M =-

0.0146M +0.0154 M =-

M = 0.01501