4P4+5S8=4P4S10
n(P4)=19.35mole
n(S8)=5.08mole
n(P4S10)=4/5*5.08*80.0%=3.25mole.
n(P4)=19.35mole
n(S8)=5.08mole
n(P4S10)=4/5*5.08*80.0%=3.25mole.
Suppose 600. G Of P4 Reacts With 1300. G Of S8. How Many Grams Of P4S10 Can Be Produced, Assuming 80.0% Yield Based On The Limiting Reactant?
