[kbakersr]

The question is:

A particular coal sample contains 3.28% S by mass. When the coal is burned, the sulfur is converted to SO₂(g). What volume of SO₂(g), measured at 23 Degrees Celsius and 738 mmHg, is produced by burning 1.2x10⁶kg of this coal?

Here was my approach:
23 Degrees Celsius = 296.15K
738 mmHg is about 738 Torr
Looking at S + O₂ --> SO₂ I figured a 1:1 ratio between S and SO₂.
so to find moles of SO₂ i found moles of S.
1.2x10⁶kg * 0.0328(the percentage) = 39360kg = 3.936x10⁷g of S
1 mol of S = 32.065g , so 3.936x10⁷g/32.065g = 1.22751x10⁶ moles of S = moles of SO₂.

so using PV = nRT, we get V=nRT/P, R is the constant 8.3145J/molK
V = ((1.22751x10⁶)*(8.3145)(296.15)) / 738 = 4.09558x10⁶

What am i doing wrong? or is that the correct answer? I don't have any way to check if the answer is correct and i feel its wrong.

[BetaAmyloid]

Check your R constant value.

You are using a R constant related to energy (J); however, you are dealing with a gas law (R constant = .08206 L*Atm/K*Mol).

[kbakersr]

Thank you

[DevaDevil]

the R is not necessarily the wrong thing here.

The only thing you want to make sure of is to use the variables in the right units.

If you want to use your 8.3145 value for R, you have to use:
p in Pascal
T in Kelvin
n in moles
and it will give you V in cubic meters.