50g of each reactant is available during reaction #2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)#, which is limiting reactant?

1 Answer
anor277
Jun 6, 2018

Would it not be the reactant with the LARGEST molecular mass..?

Explanation:

And so for...

#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#..

With respect to the reactants...we got molar quantities of...

#n_(SO_2)=(50*g)/(64*g)=0.78*mol...#

#n_(O_2)=(50*g)/(32*g)=1.56*mol...#

Dioxygen is present in stoichiometric excess, and sulfur dioxide is the LIMITING reagent....what mass of #SO_3# results given complete reaction...?

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