дано
m(Fe(OH)2) = 7.49 g
V(NH3)-?
2NH3+2H2O+FeCL2-->Fe(OH)2+2NH4CL
M(Fe(OH)2)= 90 g/mol
n(Fe(OH)2) = m/M = 7.49 / 90 = 0.08 mol
2n(NH3) = n(Fe(OH)2)
n(NH3) = 2* 0.08 = 0.16 mol
V(NH3) = Vm * n = 22.4 * 0.16 = 3.584 L
ответ 3.584 л
дано
m(Fe(OH)2) = 7.49 g
V(NH3)-?
2NH3+2H2O+FeCL2-->Fe(OH)2+2NH4CL
M(Fe(OH)2)= 90 g/mol
n(Fe(OH)2) = m/M = 7.49 / 90 = 0.08 mol
2n(NH3) = n(Fe(OH)2)
n(NH3) = 2* 0.08 = 0.16 mol
V(NH3) = Vm * n = 22.4 * 0.16 = 3.584 L
ответ 3.584 л