asked by @toriae1 •
about 1 year ago •
Chemistry
• 5 pts

In 2014, the average Kansan consumed energy equivalent to the combustion of 17,500 lbs of coal. If the chemical compositon of coal is approximated by the chemical formula C135H96O9NS, how much carbon dioxide was emitted by the combustion of this mass of coal.

Hey there! So even if you don't know the whole equation you can compute for the amount of CO2. You just need to know that you will have 13 CO2 per C135H96O9NS.

So first you need to compute for the molecular mass of coal, which I will save for you to do since I believe you can do that by yourself :)

MM C135H96O9NS = 1908.2730 g/mol

Now compute the number of moles from the given mass:

mol C135H96O9NS =(17,500 lbs/1908.2730 g/mol)(1 kg/2.2 lbs)(1000 g/1 kg) = 4168.453 mol

Now you just have to multiply by 13 to get moles of CO2:

mol CO2 = 13(4168.453 mol) = 54,189.883 mol CO2

Now get the mass if needed:

**mass CO2 = MM(mol) = (44.01 g/mol)(54,189.883 mol) = 2384896.758 g or 2384.90 kg**

I hope this helps!

answered by @leot1 •
about 1 year ago

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