дано
m(Fe3O4) = 500 g
η(Fe) = 85%
m практ (Fe)-?
3Fe3O4+8Al-->9Fe+4Al2O3
M(Fe3O4) = 232 g/mol
n(Fe3O4) = m/M = 500 / 232 = 2.16 mol
3n(Fe3O4) = 9n(Fe)
n(Fe) = 9*2.16 / 3 = 6.48 mol
M(Fe) = 56 g/mol
m теор(Fe) = n*M =6.48 * 56 = 362.88 g
m практ(Fe) = m теор(Fe) * η(Fe) /100% = 362.88 * 85% / 100% = 308.448 g
ответ 308.448 г
дано
m(Fe3O4) = 500 g
η(Fe) = 85%
m практ (Fe)-?
3Fe3O4+8Al-->9Fe+4Al2O3
M(Fe3O4) = 232 g/mol
n(Fe3O4) = m/M = 500 / 232 = 2.16 mol
3n(Fe3O4) = 9n(Fe)
n(Fe) = 9*2.16 / 3 = 6.48 mol
M(Fe) = 56 g/mol
m теор(Fe) = n*M =6.48 * 56 = 362.88 g
m практ(Fe) = m теор(Fe) * η(Fe) /100% = 362.88 * 85% / 100% = 308.448 g
ответ 308.448 г