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C2H6O = 46.08 g

508 g * (1 mol C2H6O/46.08 g C2H6O) = 11.024 mol C2H6O

There are about 11 moles in 508 grams of ethanol.

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Q: What is the number of moles in 508 g of ethanol?
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What is the total sample size (in grams) for a sample of ethanol (CH3CH2OH) which contains 3.36 g of oxygen?

9.675 g Since oxygen has an average atomic weight of 15.999 g/mol that would make it 0.21 moles of oxygen. Ethanol has one atom of oxygen per molecule so that means 0.21 moles of ethanol. Since ethanol has a molecular weight of 46.07 g/mole, 0.21 moles of ethanol would have a mass of 9.675 g.


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Unless I've completely forgotten my chemistry, mole fraction is the fraction of moles of a particular solute of the entire number of moles in the solution. So for this question: formula weight of ethanol = 46.07 g/mol formula weight of water = 18.015 g/mol moles of ethanol = 47.5g / 46.07 g/mol = 1.0310 mol moles of water = 850g / 18.015 g/mol = 47.182 mol total number of moles in the solution = 1.0310 + 47.182 = 48.213 mol therefore, the mole fraction of ethanol is 1.0310 mol / 48.213 mol = 0.0214 hope this helps.


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The formula for ethanol is CH3CH2OH So for each mole of ethanol burnt it produces 2 moles of carbon dioxide 1 Mole of ethanol is 46 g 2 moles of carbon dioxide is (2 x 44 g) 88 g So each gram of ethanol produces 1.9 grams of carbon dioxide on combustion


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