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A sample of Al²O³ contains 0.40 mol of aluminium atoms. The total mass of oxygen in the sample is:

a. 0.60 g
b. 4.3 g
c. 9.6 g
d. 19 g

Please show working out, thanks.


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1 Answer

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Best answer

n(Al) = 0.40

There are 3 O's for every 2 Al's, so n(O) = 3/2*0.40 = 0.60 mol

m(O) = 0.60 * 16 = 9.6g (Answer is C)


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