1g of ZnS combined with #O_2# to produce 0,8375g of product. What is the empirical formula of the product ?

1 Answer
anor277
Mar 4, 2018

We ASSUME that the final product is #ZnO(s)#

Explanation:

...the which results from the stoichiometric reaction....

#ZnS(s) + 3/2O_2(g) rarr ZnO(s) + SO_2(g)uarr#

And so we start with #(1.0*g)/(97.47*g*mol^-1)=0.0103*mol# with respect to zinc sulfide. Of which there were #0.0103*molxx65.4*g=0.6710*g# WITH RESPECT TO ZINC.

But we end with #0.8375*g# of zinc oxide...of which necessarily there were...#0.8375*g-0.6710*g=0.1665*g# with respect to oxygen....

And so we take the molar quantities of element to arrive at the empirical formula....

#Zn_((0.6710*g)/(65.4*g*mol^-1))O_((0.1665*g)/(16.0*g*mol^-1))-=Zn_(0.0103)O_0.0104-=ZnO# as we would anticipate....

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