Asked by Anonymous
if 7.o moles of sulfur atoms and 10 moles of oxygen molecules are combined to form the maximum amount of sulfur trioxide, how many moles of which reactant remain unused at the end?
Answers
Answered by
TutorCat
2S + 3O2 -> 2SO3
This example should help you.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
This example should help you.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
Answered by
DrBob222
2S + 3O2 ==> 2SO3
7.0 mols S x (2 mol SO3/2 mol S) = 7.0 mols SO2 formed if we had 7.0 mols S and all the O2 needed.
10 mols O2 x (2 mol SO3/3 mol O2) = 6.67 mols SO2 formed if we had 10 mols O2 and all the S needed.
In limiting reagent questions the smaller value is ALWAYS the correct one to choose; therefore, the limiting reagent is O2 and some S will be remain unreacted.
7.0 mols S x (2 mol SO3/2 mol S) = 7.0 mols SO2 formed if we had 7.0 mols S and all the O2 needed.
10 mols O2 x (2 mol SO3/3 mol O2) = 6.67 mols SO2 formed if we had 10 mols O2 and all the S needed.
In limiting reagent questions the smaller value is ALWAYS the correct one to choose; therefore, the limiting reagent is O2 and some S will be remain unreacted.
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