Oblicz, ile gramów siarczku żelaza (III) powstanie w reakcji 11,2g żelaza z siarką.
mFe=56u
mFe2S3=208u
2Fe + 3S---->Fe2S3
112g Fe------208g Fe2S3
11,2g Fe-------xg Fe2S3
x = 20,8g Fe2S3
112u 208u
2Fe + 3S --> Fe₂S₃
11,2g x
z proporcji wyliczam x
x = 20,8g
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Feci, quod potui, faciant meliora potentes
Pozdrawiam :)
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mFe=56u
mFe2S3=208u
2Fe + 3S---->Fe2S3
112g Fe------208g Fe2S3
11,2g Fe-------xg Fe2S3
x = 20,8g Fe2S3
112u 208u
2Fe + 3S --> Fe₂S₃
11,2g x
z proporcji wyliczam x
x = 20,8g
----------------------------------------------------------------------------------------------------------
Feci, quod potui, faciant meliora potentes
Pozdrawiam :)