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How many grams of Al are in 1.38 * 1023 molecules of Al2O3 if the atomic...

How many grams of Al are in 1.38 * 1023 molecules of Al2O3 if the atomic mass of Al is 27.0 g/mole, O is 16.0 g/mole, and Al2O3 is 102.0 g/mole (the final answer should have the correct number of significant figures)?  

Here is my work: (1.38*10^23 molecules) (1mol / 6.022*10^23 molecules) (27.0 gram/ mol)= 6.1873 I counted 3 Sig Fig so I rounded my answer to 6.19.... Another thought was (1.38*10^23 molecules) (54grams /1 mol) (1mol/6.022*10^23) but that answer was wrong too...

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Answer #1

According to the condition given 23 1.38 XID molecules of Al,03 have mass 23 1.38 x 10 x 102 gm 23 6.023x10 Mass 11 23.379m 1

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