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kemelbek7
@kemelbek7
August 2022
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Сколько алюминия требуется для получения 51 г оксида алюминия?
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Omejy
4Al + 3O2 --> 2Al2O3
M(Al2O3) = 102
M(Al) = 27
1) n(Al2O3) = 51/102 = 0.5моль
2)
3) m(Al) = 27*1=27г
3 votes
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Answers & Comments
M(Al2O3) = 102
M(Al) = 27
1) n(Al2O3) = 51/102 = 0.5моль
2)
3) m(Al) = 27*1=27г