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How can i Calculate the mass of oxygen in 350 g of Al(C2H3O2)3?

1 Answer

May 10, 2018

Well, what is your molar quantity of $aluminum acetate$ $"aluminum acetate"$ ?

Explanation:

You gots.... $\frac{350 \cdot g}{204.11 \cdot g \cdot m o l^{- 1}} = 1.72 \cdot m o l$ $(350*g)/(204.11 *g·mol^-1)=1.72*mol$

And given ${Al(O(O=)CCH}_{3})_{3}$ $"Al(O(O=)CCH"_3")"_3$ ....CLEARLY there are SIX moles of oxygen per mole of aluminum salt.

And so we got $10.3 \cdot m o l \cdot oxygen$ $10.3*mol*"oxygen"$ in this quantity...

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