aproned_samurai
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Report Thread starter 3 years ago
#1
Ok

I have been asked:

"how much lead (IV) oxide would be required to make up 150 grams of sodium oxide?

The equation is PbO2+2Na2SO4--> 1 Pb(SO4)2 +2 Na2O

And I am lost.

First off, I worked out the number of moles for the sodium oxide which I found to be:

150 g (mass) / 62g (gfm) =2.42 moles.

I want mass of PbO2; ergo M = N x gfm
So I figured 239.2 (gfm of
But eh that's entirely wrong

Where HAVE I went wrong......?

The Answer SHOULD be: 289 grams :\
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