Answer:
204mL
Explanation:
We'll begin by calculating the number of mole of KMnO4 in 22.5 mL of 0.374 M KMnO4. This is illustrated below:
Molarity of KMnO4 = 0.374 M
Volume = 22.5 mL = 22.5/1000 = 2.25x10^-2L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of KMnO4 = 0.374x2.25x10^-2
Mole of KMnO4 = 8.415x10^-3 mole
The equation for the reaction is given below:
10HI + 2KMnO4 + 3H2SO4 —> 5I2 + K2SO4 + 2MnSO4 + 8H2O
From the balanced equation above,
2 moles of KMnO4 reacted with 10 moles of HI.
Therefore, 8.415x10^-3 mole of KMnO4 will react with = (10x8.415x10^-3) /2 = 0.0421 mole of HI.
Now, with this amount ( i.e 0.0421 mole) of HI, we can obtain the volume of 0.206M HI solution needed for the reaction. This is illustrated below:
Molarity of HI = 0.206M
Mole of HI = 0.0421 mole
Volume =?
Molarity = mole /Volume
Volume = mole/Molarity
Volume = 0.0421/0.206
Volume = 0.204L
Now let us convert 0.204L to mL. This is illustrated below:
1L = 1000mL
0.204L = 0.204 x 1000 = 204mL
Therefore, 204mL of 0.206M HI solution is needed for the reaction.
The volume in milliliters of 0.206M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO₄ solution is 198.5mL.
Relation between molarity (M) and moles (n) will be represented as:
M = n/V, where
V = volume
Given chemical reaction is:
10HI + 2KMnO₄ + 3H₂SO₄ → 5I₂ + 2MnSO₄ + 8H₂O
Molarity of KMnO₄ = 0.372M
Volume of KMnO₄ = 22.5mL = 0.022L
Moles of KMnO₄ = 0.372M × 0.022L = 0.00818moles
From the stoichiometry of reaction it is clear that:
2 moles of KMnO₄ = react with 10 moles of HI
0.00818 moles of KMnO₄ = react with 10/2×0.00818=0.0409 moles of HI
Given molarity of HI = 0.206M
Now we calculate the volume of HI by using below formula:
V = n/M
V = 0.0409moles / 0.206M = 0.1985 L = 198.5mL
Hence, volume in mL is 198.5mL.
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https://brainly.com/question/26968296
