дано
m(FeCL3) = 234 g
m(Fe)-?
2Fe+3CL2-->2FeCL3
M(FeCL3) = 162.5 g/mol
n(FeCL3) = m/M = 234 / 162.5 = 1.44 mol
2n(Fe) = 2n(FeCL3) = 1.44 mol
M(Fe) = 56 g/mol
m(Fe) = n*M = 1.44 * 56 = 80.64 g
ответ 80.64 г
дано
m(FeCL3) = 234 g
m(Fe)-?
2Fe+3CL2-->2FeCL3
M(FeCL3) = 162.5 g/mol
n(FeCL3) = m/M = 234 / 162.5 = 1.44 mol
2n(Fe) = 2n(FeCL3) = 1.44 mol
M(Fe) = 56 g/mol
m(Fe) = n*M = 1.44 * 56 = 80.64 g
ответ 80.64 г