4.19 g of potassium permanganate, KMnO4, are mixed with excess magnesium iodide. a. Write a balanced chemical equation that corresponds to this reaction. question a,b, and c

Question

1.In a double replacement reaction 4.19 g of potassium permanganate, KMnO4, are mixed with excess magnesium iodide.

a. Write a balanced chemical equation that corresponds to this reaction.

b. Determine the mass of magnesium permanganate formed.

c. If the actual yield of magnesium permanganate was 2.92 g, determine percent yield of this reaction.

Answer 1

2K(MnO4) + MgI2 → 2KI + Mg(MnO4)2 Balanced equation

Calculate Molar Mass:

KMnO4=39.1amu+119amu=158g/mol

Mg(MnO4)2=24.3amu+119amu=262g/mol

Calculate grams of Mg(MnO4)2

4.19gKMnO4 x 1molKMnO4/158gKMnO4 x 1molMg(MnO4)2/2molKMnO4 x 262gMg(MnO4)/molMg(MnO4)2

= 34.7gMg(MnO4)2

Calculate percent yield:

2.92gMg(MnO4)2/3.47gMg(MnO4)2x100%=84.1%