Question

Какой объём кислорода при н.у. потребуется для сжигания серы массой 72 г ?

Answer 1

Ответ:

в фото

Объяснение:

так как не указан какой оксид получится в продуктах реакции, рассчитаю объем кислорода для SO2

Answer 2

Ответ:

для полного сгорания серы будет нужно 5,6л кислорода.

Answer 3

Calculating Oxygen Volume Required for Sulfur Combustion at STP

To calculate the volume of oxygen required for the combustion of sulfur at standard temperature and pressure (STP), we can use the balanced chemical equation for the reaction between sulfur and oxygen.

The balanced chemical equation for the combustion of sulfur is: S(s) + O2(g) → SO2(g)

From the equation, we can see that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide.

Given that the mass of sulfur is 72 g, we can calculate the number of moles of sulfur using its molar mass.

The molar mass of sulfur (S) is approximately 32.07 g/mol.

Using the formula: Number of moles = Mass / Molar mass

We can calculate the number of moles of sulfur: Number of moles = 72 g / 32.07 g/mol

Number of moles of sulfur ≈ 2.24 moles

Since the balanced chemical equation shows that 1 mole of sulfur reacts with 1 mole of oxygen, we need the same number of moles of oxygen as moles of sulfur for complete combustion.

Now, to find the volume of oxygen at STP, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

Volume of oxygen required = Number of moles of sulfur × Molar volume at STP

Volume of oxygen required ≈ 2.24 moles × 22.4 L/mol

Volume of oxygen required ≈ 50.18 L

Therefore, approximately 50.18 liters of oxygen at STP would be required for the combustion of 72 grams of sulfur.