Oblicz masę bromometanu powstałego w procesie bromowania 3 moli metanu w obecności światła. Dokładne obliczenia.
mCH3Br=95u
CH4 + Br2---UV--->CH3Br + HBr
1mol CH4-------95g CH3Br
3mole CH4-------xg CH3Br
x = 3*95
x = 285g CH3Br
3CH4 + 3Br2 -> 3CH3Br + 3HBr
16g_____________95g
3x16g___________X g
X = 3x16g x 95g / 16g
X = 285g - masa 3 moli bromometanu
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mCH3Br=95u
CH4 + Br2---UV--->CH3Br + HBr
1mol CH4-------95g CH3Br
3mole CH4-------xg CH3Br
x = 3*95
x = 285g CH3Br
3CH4 + 3Br2 -> 3CH3Br + 3HBr
16g_____________95g
3x16g___________X g
X = 3x16g x 95g / 16g
X = 285g - masa 3 moli bromometanu