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18g --- 6,02 * 10²³ at. O
36,04g --- x
x = 12,053 * 10²³ atomów tlenu
1 mol Al₂O₃ ma masę 102 g/mol i znajduje się tam 3 * 6,02 * 10²³ atomów tlenu
102g --- 3 * 6,02 * 10²³ at. O
x --- 12,053 * 10²³ at. O
x = 68,07 g Al₂O₃