дано
m(ppa FeCL3) = 56 g
W(FeCL3) = 5%
+NaOH
m(FE(OH)3)-?
m(FeCL3) = 56 * 5% / 100% = 2,8 g
FeCL3+3NaOH-->FE(OH)3+3NaCL
M(FECL3) = 162.5 g/mol
n(FeCL3) = m/M = 2.8 / 162.5 = 0.017 mol
n(FeCL3) = n(Fe(OH)3)
M(Fe(OH)3) = 107 g/mol
m(Fe(OH)3) = n*M = 0.017*107 = 1.819 g
ответ 1.819 г
Объяснение:
дано
m(ppa FeCL3) = 56 g
W(FeCL3) = 5%
+NaOH
m(FE(OH)3)-?
m(FeCL3) = 56 * 5% / 100% = 2,8 g
FeCL3+3NaOH-->FE(OH)3+3NaCL
M(FECL3) = 162.5 g/mol
n(FeCL3) = m/M = 2.8 / 162.5 = 0.017 mol
n(FeCL3) = n(Fe(OH)3)
M(Fe(OH)3) = 107 g/mol
m(Fe(OH)3) = n*M = 0.017*107 = 1.819 g
ответ 1.819 г
Объяснение: