Question

How many grams of water would be produced by the decompostion of 1.07 grams of H2O2?

Answer 1

Well, we need a stoichiometric equation....

Explanation:

And so...

`H_2O_2(g) stackrel"catalysis"rarr H_2O(l)+1/2O_2(g)uarr`

`"Moles of hydrogen peroxide"=(1.07*g)/(34.01*g*mol^-1)=0.0315*mol`.

And thus we generate an half equiv dioxygen gas...and ONE equiv of water...

`0.0315*molxx18.01*g*mol^-1=0.568*g`...

The catalyst is usually a `Mn(IV+)` salt, i.e. `MnO_2`...