дано
m(ppaNaCL) = 292.5 g
W(NaCL) = 10%
m(ppa AgNO3) = 136 g
W(AgNO3) = 25%
m(AgCL) - ?
NaCL+AgNO3-->AgCL+NaNO3
m(NaCL) = m(ppa NaCL) * W(NaCL) / 100% = 292.5 *10 / 100 = 29.25 g
m(AgNO3) = m(ppa AgNO3) * W(AgNO3) / 100% = 136 * 25 / 100 = 34 g
M(NaCL) = 58.5 g/mol
n(NaCL) = m(NaCL) / M(NaCL) = 29.25 / 58.5 = 0.5 mol
M(AgNO3) = 170 g/mol
n(AgNO3) = m(AgNO3) / M(AgNO3) = 34 / 170 = 0.2 mol
n(NaCL) > n(AgNO3)
n(AgNO3) = n(AgCL) = 0.2 mol
M(AgCL) = 143,5 g/mol
m(AgCL) = n(AgCL) * M(AgCL) = 0.2*143.5 = 28.7 g
ответ 28.7 гр
Объяснение:
M(C6H5OK)=132 кг/кмоль
n(C6H5OK)=10/132=0.076 кмоль
n(C6H5OH)=n(C6H5OK)=0,076 кмоль
M(C6H5OH)=94 г/моль
m(C6H5OH)=94х0,076=7,144 кг
n(H2SO4)=0,076/2=0,038 кмоль
M(H2SO4)=98 кг/кмоль
m(H2SO4)=98х0,038=3,724 кг
m(р-ра H2SO4)=3,724/0,9=4,138 кг
V(р-ра H2SO4)=4,138/1,8=2,3 л