Question

Какая масса перманганата калия потребуется для получения 11.2 л кислорода н.у

Answer 1

Ответ:

2KMnO4(1моль)-->K2MnO4+MnO2+O2(0,5моль)

Объяснение:

n(O2)=V(O2):Vm(O2)=11,2:22,4=0,5 моль

m(KMnO4)=n(KMnO4)×M(KMnO4)=1×158=158г

Answer 2

Calculating the Mass of Potassium Permanganate Required

To calculate the mass of potassium permanganate required to obtain 11.2 L of oxygen at standard conditions, we can use the balanced chemical equation for the decomposition of potassium permanganate.

The balanced chemical equation for the decomposition of potassium permanganate is: 2 KMnO4 → K2MnO4 + MnO2 + O2

From the balanced equation, we can see that 2 moles of potassium permanganate produce 1 mole of oxygen gas.

Given that the molar volume of a gas at standard conditions (0°C and 1 atm) is approximately 22.4 L, we can calculate the moles of oxygen gas produced and then use the molar mass of potassium permanganate to find the mass required.

Calculation Steps

1. Calculate the moles of oxygen gas required: - Moles of O2 = Volume of O2 / Molar volume at STP - Moles of O2 = 11.2 L / 22.4 L/mol - Moles of O2 = 0.5 moles

2. Use the mole ratio from the balanced equation to find the moles of potassium permanganate required: - Moles of KMnO4 = 0.5 moles O2 * (2 moles KMnO4 / 1 mole O2) - Moles of KMnO4 = 1 mole

3. Calculate the mass of potassium permanganate required: - Mass of KMnO4 = Moles of KMnO4 * Molar mass of KMnO4

Molar Mass of Potassium Permanganate

The molar mass of potassium permanganate (KMnO4) is approximately 158.034 g/mol.

Final Calculation

- Mass of KMnO4 = 1 mole * 158.034 g/mol - Mass of KMnO4 = 158.034 grams

Therefore, to obtain 11.2 L of oxygen at standard conditions, approximately 158.034 grams of potassium permanganate would be required.