Question

Какой объём кислорода потребуется при сжигании 320 гр серы?

Answer 1

S + O2 =SO2

 

кол-во серы = 320/32=10 моль

кол-во ксилорода = кол-во серы = 10 моль

объем кислорода = 10*22,4=224л

Answer 2

Calculating Oxygen Consumption in Sulfur Combustion

When sulfur is burned, it reacts with oxygen to produce sulfur dioxide. To calculate the amount of oxygen required for the combustion of 320 grams of sulfur, we can use the balanced chemical equation for the reaction:

S(s) + O2(g) → SO2(g)

The balanced equation shows that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide.

Calculation Steps

1. Convert grams of sulfur to moles: - The molar mass of sulfur (S) is approximately 32.06 g/mol. - Therefore, 320 grams of sulfur is equal to 320 g / 32.06 g/mol = 9.97 moles of sulfur.

2. Determine the moles of oxygen required: - From the balanced chemical equation, 1 mole of sulfur requires 1 mole of oxygen. - Therefore, 9.97 moles of sulfur will require 9.97 moles of oxygen.

3. Convert moles of oxygen to volume: - At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. - Therefore, the volume of oxygen required for the combustion of 320 grams of sulfur is 9.97 moles * 22.4 L/mole = 223.63 liters.

Conclusion

The combustion of 320 grams of sulfur will require approximately 223.63 liters of oxygen.

This calculation is based on the stoichiometry of the chemical reaction and the principles of ideal gas behavior.