How many grams of #O_2# are required to produce 358.5 grams of #ZnO# in #2Zn + O_2 -> 2ZnO#?

1 Answer
anor277
Mar 9, 2017

We have the stoichiometric equation.........

Explanation:

#Zn(s) + 1/2O_2(g) rarr ZnO(s)#

And we have #(358.5*g)/(81.41*g*mol^-1)=4.40*mol# of zinc oxide.

Given the stoichiometry, #2.20*mol# of dioxygen gas were necessary, i.e. #2.20*molxx32.00*g*mol^-1=70.0*g#

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