Question (1 Viewer)

[steph@nie]

I don't know if this is even in the right module, but anyway.

Coal containing 1% sulfur, is burned in a power station.

Calculate the volume of sulfur dioxide released at 25 C and 100kPa when 10.0 million kg of coal is burnt.

 

[Xayma]

Im assuming 1% sulfur is w/w.

∴ you are combusting 100 000kg of Sulfur.

ie you are combusting 3.12Mmol
Now 1 mole of Sulfur produces 1 mole of Sulfur Dioxide.

∴ 3.12Mmol*24.79
=77.34ML

 

[Tommy_Lamp]

yeah like Xayma said, all its asking you to do is determine 1% of 10mil kg. thats the amount of sulfur. Then you have
S + O2 --> SO2
m/mm = v/mv
100,000,000/32.06 = v/24.47
v = 76,325,639L
v = 76ML

hmm xayma, different answer

EDIT: my bad

 

[steph@nie]

different molar volumes... I would've used the data sheet value.. but thankyou to both of you.

edit: they wanted 24.79

 

[Not Too Bright]

that reminds me the answer for question 23 in the 2004 CSSA is wrong it should be about 7.7 million litres

 

[Paroissien]

that reminds me the answer for question 23 in the 2004 CSSA is wrong it should be about 7.7 million litres

Thanks for that. I was almost positive it was, but this just confirms it.
And this isn't that question exactly. That was 0.1% sulfur in the coal.

 

[smallcattle]

what about 2003 HSC Q22 b

where it wants volume of CO2 when 72.5g of ethanol was burnt completely?

 

[Paroissien]

what about 2003 HSC Q22 b

where it wants volume of CO2 when 72.5g of ethanol was burnt completely?

Firstly I always like to start with an equation:
CH5OH + 3O2 ---> 2CO2 + 3H2O
moles (ethanol) = mass/molar mass
= 72.5/46 = 1.576 moles
moles (CO2) = 2*1.576
= 3.152 moles of CO2
volume (CO2) = moles*24.79
= 78.1 L of CO2