Respuesta :

meerkat18
There is an exact value for the standard volume at standard conditions of 1 atm and 273 K. This standard volume for any ideal gas is 22.4 L/mol. Thus,

Moles SO₂ = 5.9 L * 1 mol/22.4 L = 0.263 mol

The molar mass for SO₂ is 64.066 g/mol. So, the mass is:

Mass = 0.263 mol * 64.066 g/mol = 16.87 g SO₂
Eduard22sly

16.83 g of SO₂ is present in the 5.9 L container at stp.

We'll begin by calculating the number of mole of SO₂ present in the 5.9 L container.

At standard temperature and pressure (stp),

22.4 L = 1 mole of SO₂

Therefore,

5.9 L = 5.9 / 22.4

5.9 L = 0.263 mole of SO₂

Thus, 0.263 mole of SO₂ is present in the container.

Finally, we shall determine the mass of 0.263 mole of SO₂. This can be obtained as follow:

Mole of SO₂ = 0.263

Molar mass of SO₂ = 32 + (16×2)

= 32 + 32

= 64 g/mol

Mass of SO₂ =?

Mass = mole × molar mass

Mass of SO₂ = 0.263 × 64

Mass of SO₂ = 16.83 g

Therefore, 16.83 g of SO₂ is present in the 5.9 L container at stp.

Learn more: https://brainly.com/question/9945437

Otras preguntas