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mcz=27u+3*14u+6*16u=165u
Al= 27u
N₂= 42u
O₂= 96u
27u:42u:96u /:3u
9:14:32
Stosunek masowy Al(NO₂)₃ wynosi 9:14:32
P.S Wiem, zagapiłem się ;)