Ответ:
дано
m(C6H5CL) = 128 g
m пр (C6H5OH) = 120 g
--------------------
η(C6H5OH) - ?
C6H5CL+NaOH-->C6H5OH+NaCL
M(C6H5CL) = 112.5 g/mol
M(C6H5OH) = 94 g/mol
m(C6H5CL) / M(C6H5CL) = m теор (C6H5OH) / M(C6H5OH)
m теор(C6H5OH) = 128 * 94 / 112.5 = 107 g
η(C6H5OH) = m теор(C6H5OH) /m пр (C6H5OH) * 100%
η(C6H5OH) = 107 / 120 * 100% = 89.16%
ответ 89.16%
Объяснение:
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Ответ:
дано
m(C6H5CL) = 128 g
m пр (C6H5OH) = 120 g
--------------------
η(C6H5OH) - ?
C6H5CL+NaOH-->C6H5OH+NaCL
M(C6H5CL) = 112.5 g/mol
M(C6H5OH) = 94 g/mol
m(C6H5CL) / M(C6H5CL) = m теор (C6H5OH) / M(C6H5OH)
m теор(C6H5OH) = 128 * 94 / 112.5 = 107 g
η(C6H5OH) = m теор(C6H5OH) /m пр (C6H5OH) * 100%
η(C6H5OH) = 107 / 120 * 100% = 89.16%
ответ 89.16%
Объяснение: