Question

I'm trying to find out how many grams of water can be produced in a reaction of 11.7 moles of ethane with oxygen gas. The balanced equation is 2C2H6 + 7O2 yields 4CO2 + 6H2O. I need help determining the setup of the conversion. I think it would be 11.7 mol C2H6 x 54.039 mol H2O/1 mol C2H6 x something over 3 mol H2O...help? :)

Answer 1

@Australopithecus

Answer 2

@waxynaveed

Answer 3

from balanced equation,
2 moles of ethane can produce= 6 moles of water
11.7 moles . .. .. .. .. . . . .. . = 6/2 X 11.7
...................................................... = 35.1 moles of water

mass of water = moles X m.mass of H2O
mass = 35.1 X 18 = 631.8 g of water will produce when 11.7 moles of ethane will react with oxygen

Answer 4

So this is simple,

2C2H6 + 7O2 -> 4CO2 + 6H2O

First off your limiting reagent is the ethanol 11.7 mole you assume oxygen in excess in this question

so for every 2 moles of ethanol reacted 6 moles of water are produced that is what the equation is telling me,
so divide by 2, then multiply by 6 to find the moles of water produced, then simply use the equation,

moles = grams/molecular mass

in this case you want the molecular mass of water

Answer 5

Where did you get the 18 from???

Answer 6

18 is the molecular mass of water, he should have put units grams/mole

Answer 7

15.9994g/mol for the molecular mass of oxygen
1.0079g/mol for the molecular mass of hydrogen

he just rounded them,

so,

H2O consists of two hydrogens and one oxygen so,

1.0g/mol + 1.0g/mol + 16g/mol = 18g/mol molecular mass

Answer 8

H2O= 2+16= 18
molar mass of water

Answer 9

roughly the molecular mass of water