Ответ:
9.2г Al4C3
----------------------
дано
V(CH4) = 4.3 L
-------------------
m(Al4C3) - ?
Al4C3+12H2O-->4Al(OH)3+3CH4
n(CH4) = V(CH4) / Vm = 4.3 / 22.4 = 0.2 mol
n(Al4C3) = 3n(CH4)
n(Al4C3) = 0.2 / 3 = 0.067 mol
M(Al4C3) = 144 g/mol
m(Al4C3) = n*M = 0.067 * 144 = 9.648 g
ответ 9.648 гр
Объяснение:
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Ответ:
9.2г Al4C3
----------------------
Ответ:
дано
V(CH4) = 4.3 L
-------------------
m(Al4C3) - ?
Al4C3+12H2O-->4Al(OH)3+3CH4
n(CH4) = V(CH4) / Vm = 4.3 / 22.4 = 0.2 mol
n(Al4C3) = 3n(CH4)
n(Al4C3) = 0.2 / 3 = 0.067 mol
M(Al4C3) = 144 g/mol
m(Al4C3) = n*M = 0.067 * 144 = 9.648 g
ответ 9.648 гр
Объяснение: