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Вычислите массовые доли элементов в следующих соединениях:
а) Al2O3; б) P2S5; в) Fe3O4; г) Al4C3.
а)
Ответ: `ω(Al) = 52.9%`, `ω(O) = 47.1%`. |
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| Дано: | Решение |
|---|---|
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`Al_2O_3` |
`ω(Al) = (100*k*A_r(Al))/(M_r(Al_2O_3)) = (100*2*27)/102 = 52.9%` `ω(O) = (100*k*A_r(O))/(M_r(Al_2O_3)) = (100*3*16)/102 = 47.1%` |
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`ω(Al) = ?` `ω(O) = ?` |
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б)
Ответ: `ω(P) = 27.9%`, `ω(S) = 72.1%`. |
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| Дано: | Решение |
|---|---|
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`P_2S_5` |
`ω(P) = (100*k*A_r(P))/(M_r(P_2S_5)) = (100*2*31)/222 = 27.9%` `ω(S) = (100*k*A_r(S))/(M_r(P_2S_5)) = (100*5*32)/222 = 72.1%` |
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`ω(P) = ?` `ω(S) = ?` |
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в)
Ответ: `ω(Fe) = 72.4%`, `ω(O) = 27.6%`. |
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| Дано: | Решение |
|---|---|
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`Fe_3O_4` |
`ω(Fe) = (100*k*A_r(Fe))/(M_r(Fe_3O_4)) = (100*3*56)/232 = 72.4%` `ω(O) = (100*k*A_r(O))/(M_r(Fe_3O_4)) = (100*4*16)/232 = 27.6%` |
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`ω(Fe) = ?` `ω(O) = ?` |
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г)
Ответ: `ω(Al) = 75%`, `ω(C) = 25%`. |
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| Дано: | Решение |
|---|---|
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`Al_4C_3` |
`ω(Al) = (100*k*A_r(Al))/(M_r(Al_4C_3)) = (100*4*27)/144 = 75%` `ω(C) = (100*k*A_r(C))/(M_r(Al_4C_3)) = (100*3*12)/144 = 25%` |
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`ω(Al) = ?` `ω(C) = ?` |
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