Question

Определить массу кислорода,который образуется при разложении 5 моль KMnO4---->K2MnO4+MnO2+O2

Answer 1

Уравнение реакции:
2KMnO4 --(t°)--> K2MnO4 + MnO2 + O2
v(KMnO4) : v(O2) = 2 : 1 => v(O2) = 2,5 моль.
m(O2) = 2,5 моль * 32 г/моль = 80 г.

Answer 2

2KMnO4 -----> K2MnO4 + MnO2 + O2
n(KMnO4) = 5 моль / 158 г / моль = 0,03
n(O2) = 0,03 / 2 = 0,015 моль
m(O2) = 0,015 моль * 32 г / моль = 0,96 г

Answer 3

Calculating the Mass of Oxygen Produced

To determine the mass of oxygen produced during the decomposition of 5 moles of KMnO4, we can use stoichiometry to find the molar ratio of KMnO4 to O2 and then calculate the mass of oxygen produced.

The balanced chemical equation for the decomposition of KMnO4 is: 5 KMnO4 → K2MnO4 + MnO2 + 3 O2

Using Stoichiometry to Find the Molar Ratio

The molar ratio between KMnO4 and O2 in the balanced chemical equation is 5:3, which means that for every 5 moles of KMnO4, 3 moles of O2 are produced.

Calculating the Mass of Oxygen Produced

To calculate the mass of oxygen produced, we can use the molar mass of O2, which is approximately 32 g/mol.

Using the molar ratio and the molar mass of O2, we can calculate the mass of oxygen produced as follows: - 5 moles of KMnO4 produce (3/5) * 5 moles of O2 - The mass of (3/5) * 5 moles of O2 is (3/5) * 5 * 32 g

Final Calculation

Therefore, the mass of oxygen produced when 5 moles of KMnO4 decompose is: ((3/5) * 5 * 32) g

Result

The mass of oxygen produced when 5 moles of KMnO4 decompose is 96 g.