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

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Answer 1 · 2017-03-09T22:05:02

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

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

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. $2.20*molxx32.00*g*mol^-1=70.0*g$

How many grams of O_2O_2 are required to produce 358.5 grams of ZnOZnO in 2Zn + O_2 -> 2ZnO2Zn + O_2 -> 2ZnO?