Question
Asked Sep 29, 2019
58. What volume of 0.0665 M KMnO4 is necessary to con
vert 12.5 g KI to Ik in the reaction below? Assume that
H_SO, is present in excess.
2 KMnO410 KI + 8 H2SO4
6 K2SO2 MnSO4 +5 28 H0
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58. What volume of 0.0665 M KMnO4 is necessary to con vert 12.5 g KI to Ik in the reaction below? Assume that H_SO, is present in excess. 2 KMnO410 KI + 8 H2SO4 6 K2SO2 MnSO4 +5 28 H0

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Step 1

The given balanced chemical reaction is given as:

2 KMnO4 + 10 KI + 8 H2SO4 → 6 K2SO4 + 2 MnSO4 + 5 I2 + 8 H2O

 

Also given,

Molarity of KMnO4 = 0.0665 M = 0.0665 mol/L

Mass of KI needed to produce I2 = 12.5 g

Step 2

Moles of KI needed in this reaction can be calculated as:

Molar mass of KI = 39 + 127 166 g/mol
12.5 g
Mass
0.075 mol
Moles Molar mass 166 g/mol
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Molar mass of KI = 39 + 127 166 g/mol 12.5 g Mass 0.075 mol Moles Molar mass 166 g/mol

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Step 3

From the given reaction, it is evident that 10 mol of KI reacts with 2 mol of KMnO4 to produce I2. Therefo...

10 mol of KI reacts with
2 mol of KMnO4
2 mol
.0.075 mol of KI reacts with
x 0.075 mol
10 mol
- 0.015 mol of KMnO4
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10 mol of KI reacts with 2 mol of KMnO4 2 mol .0.075 mol of KI reacts with x 0.075 mol 10 mol - 0.015 mol of KMnO4

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