дано
m(ppaC6H5OH) = 100 g
W(C6H5OH) = 5%
+KJ
m(C6H5OK)-?
m(C6H5OH) = 100*5% / 100% = 5 g
C6H5OH+KJ-->C6H5OK+HJ
M(C6H5OH) = 94 g/mol
n(C6H5OH) = m/M =5/ 94 = 0.053 mol
n(C6H5OH) = n(C6H5OK) = 0.053 mol
M(C6H5OK)=132 g/mol
m(C6H5OK) = n*M = 0.053 * 132 = 7g
ответ 7 г
дано
m(ppaC6H5OH) = 100 g
W(C6H5OH) = 5%
+KJ
m(C6H5OK)-?
m(C6H5OH) = 100*5% / 100% = 5 g
C6H5OH+KJ-->C6H5OK+HJ
M(C6H5OH) = 94 g/mol
n(C6H5OH) = m/M =5/ 94 = 0.053 mol
n(C6H5OH) = n(C6H5OK) = 0.053 mol
M(C6H5OK)=132 g/mol
m(C6H5OK) = n*M = 0.053 * 132 = 7g
ответ 7 г