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stiks02 [169]
3 years ago
5

A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o

xidize the mn2+ to mno4 - ?
Chemistry
1 answer:
azamat3 years ago
5 0
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
                                               ≈ 31.4 mg
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<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

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