1
дано
m(ppa C6H5OH) = 235 g
W(C6H5OH)= 10%
m(Br2)-?
m(C6H2Br3OH)-?
m(C6H5OH)= 235*10% / 100% = 23.5 g
23.5 X y
C6H5OH+3Br2-->C6H2Br3OH+3HBr
94 3*160 331
M(C6H5OH) = 94 g/mol M(Br2)= 160 g/mol
M(C6H2Br3OH) = 331 g/mol
X = 23.5*480 / 94 = 120 g
y = 23.5 * 331 / 94 = 82.75 g
ответ 120 г, 82.75 g
2.
дано
m(практ C6H5OH) = 4.7 g
η(C6H5OH) = 70%
m(C6H5CL)-?
m(C6H5OH) = 4.7*100% / 70% = 6.7 g
X 6.7
C6H5CL+NaOH-->C6H5OH+NaCL
112.5 94
M(C6H5CL) = 112.5 g/mol M(C6H5OH) = 94 g/mol
X= 112.5*6.7 / 94 = 8 g
ответ 8 г
3. дано
m(C6H2O(NO2)3) =2.29 g
m(C6H5OH)-?
m(HNO3)-?
X 2.29
C6H5OH+3HNO3-->C6H2O(NO2)3+3H2O
3*63 228
X = 3*63 * 2.29 / 228 = 1.9 g
M(HNO3) = 63 g/mol
M(C6H2O(NO2)3) = 228 g/mol
n(C6H2O(NO2)3) = m/M = 2.29 / 228 = 0.01 mol
n(C6H5O(NO2)3)= n(C6H5OH) = 0.01 mol
M(C6H5OH) = 94 g/mol
m(C6H5OH) = n*M = 0.01*94 = 0.94 g
ответ m(HNO3)= 1.9 g, m(C6H5OH) = 0.94 g
W(Fe) = Ar(Fe) * n / Mr(Fe2O3) *100% = 56 * 2 / 160 *100% = 70%
W(O) = Ar(O) * n / Mr(Fe2O3) *100% = 16 * 3 / 160 *100% = 30%