Question

Сколько граммов йода выделится при окислении раствора йодида натрия 150 мл 0.02 М раствора перманганат калия в кислой среде?

Answer 1

Дано:V(KMnO4) = 150 млρ(KMnO4) = 1,08 г\млw(KMnO4) = 6%Найти:m(J2)Решение:Вычислим массу раствора перманганата калия:m (р-ра KMnO4)  = ρ(р-ра KMnO4) ∙ V(р-ра KMnO4)  =  1,08г\ мл ∙  150 мл = 162 гВычислим массу KMnO4 в растворе:m(KMnO4) = m(р-ра KMnO4) ∙  w(р-ра KMnO4)  \ 100%m(KMnO4) = 162 г ∙  6% \ 100% = 9,72 гНайдем объем выделившегося иода:                                               9,72 г            х г10KJ + 8H2SO4 (разб.) + 2KMnO4 = 5J2 ↓ + 2MnSO4 + 8H2O + 6K2SO4.                                               316 г           1270 г 9,72 г -- х г316 г -- 1270 гх = 9,72 ∙ 1270 г\ 316 г = 39 гОтвет: m(J2) = 39 г

Answer 2

Calculating the Amount of Iodine Released

To calculate the amount of iodine (I2) that will be released when a 150 mL 0.02 M sodium iodide (NaI) solution is oxidized by a solution of potassium permanganate (KMnO4) in acidic medium, we can use the following balanced chemical equation:

5NaI + 2KMnO4 + 8H2SO4 → 5Na2SO4 + 2MnSO4 + 8H2O + 2K2SO4 + 5I2

The molar ratio between NaI and I2 is 5:5, which simplifies to 1:1. This means that for every mole of NaI, one mole of I2 will be produced.

Given that the volume of the NaI solution is 150 mL and the concentration is 0.02 M, we can calculate the number of moles of NaI present in the solution:

Number of moles of NaI = Volume (L) × Concentration (M) Number of moles of NaI = 0.150 L × 0.02 mol/L Number of moles of NaI = 0.003 moles

Since the molar ratio between NaI and I2 is 1:1, the number of moles of I2 produced will be the same as the number of moles of NaI.

Now, we can calculate the mass of I2 produced using the molar mass of I2, which is 253.8 g/mol.

Mass of I2 = Number of moles of I2 × Molar mass of I2 Mass of I2 = 0.003 moles × 253.8 g/mol Mass of I2 = 0.7614 grams

Therefore, approximately 0.7614 grams of iodine will be released when the given reaction occurs.