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Question

C2H5OH + 3 O2 > 2 CO2 + 3 H2O + 1570 kJ

Calculate the mass in grams of carbon dioxide that would be produced along with the release of 2.71x10^5 kJ.

Your help is greatly appreciated...

Answer 1

(2.71x10^5 kJ/1570)x2=number of molecules of CO2. So, how many molecules of CO2 are in a gram? We know that the molecular weight of CO2 is 44.01, so 1 mole of CO2 will weigh 44.01grams. We should also know that 1 mole contains avagadros number of molecules (6.023 X 10^23), so 1 gram CO2 is avagodros number/44.01. Now you can divide the number of molecules of CO2 you produced by the number of molecules in a gram to give you the mass in grams.

Answer 2

Well, the equaiton tells us that there will be 2 moles of CO2 released for every 1570 kJ released.

so, 2.71*10^5/1570 gives us the factor for how many times more energy was released than for one equation worth
=172.6

therefore, 172.6*2 gives the number of moles of CO2 produced with the described total energy release

345.2 moles of CO2 weighs 44.01 gm/mole

and that means, 345.2*44.01=15193.3 gms
or, 15.2 kg

you should probably check my arithmetic

Answer 3

dunno