дано
m(Al4C3) = 54 g
V(CH4)-?
Al4C3+12H2O-->4Al(OH)3+3CH4
M(Al4C3) = 144 g/mol
n( Al4C3) = m/M = 54 / 144 = 0.375 mol
n(Al4C3) = 3n(CH4)
n(CH4)= 3*0.375 = 1.125 mol
V(CH4) = n*Vm = 1.125 * 22.4 = 25.2 L
ответ 25.2 л
Объяснение:
дано
m(Al4C3) = 54 g
V(CH4)-?
Al4C3+12H2O-->4Al(OH)3+3CH4
M(Al4C3) = 144 g/mol
n( Al4C3) = m/M = 54 / 144 = 0.375 mol
n(Al4C3) = 3n(CH4)
n(CH4)= 3*0.375 = 1.125 mol
V(CH4) = n*Vm = 1.125 * 22.4 = 25.2 L
ответ 25.2 л
Объяснение: