When coal is burned, the sulfur present in coal is converted to sulfur dioxide (SO2), which is responsible for the acid rain phenomenon. S(s) + O2(g) → SO2(g) If 4.10 kg of S are reacted with oxygen, calculate the volume of SO2 gas (in mL) formed at 30°C and 1.16 atm.

Respuesta :

Answer : The volume of [tex]SO_2[/tex] gas is, [tex]2.75\times 10^6mL[/tex]

Explanation : Given,

Mass of S = 4.10 kg = 4100 g      (1 kg = 1000 g)

Molar mass of S = 32 g/mol

First we have to calculate the moles of S.

[tex]\text{Moles of }S=\frac{\text{Given mass }S}{\text{Molar mass }S}=\frac{4100g}{32g/mol}=128.125mol[/tex]

Now we have to calculate the moles of [tex]SO_2[/tex]

The balanced chemical reaction is:

[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]

From the balanced reaction, we conclude that

As, 1 mole of [tex]S[/tex] react to give 1 mole of [tex]SO_2[/tex]

So, 128.125 mole of [tex]S[/tex] react to give 128.125 mole of [tex]SO_2[/tex]

Now we have to calculate the volume of [tex]SO_2[/tex] by using ideal gas equation.

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]SO_2[/tex] gas = 1.16 atm

V = Volume of [tex]SO_2[/tex] gas = ?

n = number of moles [tex]SO_2[/tex] = 128.125 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]SO_2[/tex] gas = [tex]30^oC=273+30=303K[/tex]

Putting values in above equation, we get:

[tex]1.16atm\times V=128.125mole\times (0.0821L.atm/mol.K)\times 303K[/tex]

[tex]V=2747.65L=2.75\times 10^6mL[/tex]     (1 L = 1000 mL)

Therefore, the volume of [tex]SO_2[/tex] gas is, [tex]2.75\times 10^6mL[/tex]

The volume of SO2 gas (in mL) formed at 30°C and 1.16 atm should be 2.75*10^6 mL.

Calculation of the volume:

Since

Mass of S = 4.10 kg = 4100 g      (1 kg = 1000 g)

Molar mass of S = 32 g/mol

Here first determine the mole of S

So,

= 4100 / 32

= 128.125 mol

Now the volume should be

We know that

PV = nRT

Here

P = Pressure of  gas = 1.16 atm

n = number of moles  = 128.125 mole

R = Gas constant = 0.0821 L

T = Temperature of  gas = 30 degrees = 273 + 30 = 303 k

So,

1.16atm * v = 128.125 * (0.0821) * 303

= 2747.65 L

= 2.75*10^6 mL

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