дано
m(Fe) = 38 g
η(H2) - 85%
V(H2)-?
Fe+2HCL-->FeCL2+H2
M(Fe) = 56 g/mol
n(Fe) = m/M = 38 / 56 = 0.68 mol
n(Fe) = n(H2) = 0.68 mol
V теор(H2) = n*Vm = 0.68 * 22.4 = 15.232 l
Vпрак(H2) = 15.232 * 85% / 100% = 13 l
ответ 13 л
Объяснение:
дано
m(Fe) = 38 g
η(H2) - 85%
V(H2)-?
Fe+2HCL-->FeCL2+H2
M(Fe) = 56 g/mol
n(Fe) = m/M = 38 / 56 = 0.68 mol
n(Fe) = n(H2) = 0.68 mol
V теор(H2) = n*Vm = 0.68 * 22.4 = 15.232 l
Vпрак(H2) = 15.232 * 85% / 100% = 13 l
ответ 13 л
Объяснение: