дано
m практ (C6H5ONa) = 15.08 g
m(C6H5CL) = 22.5 g
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η(C6H5ONa)-?
C6H5CL+2NaOH--->C6H5ONa+NaCL+H2O
M(C6H5CL) = 112.5 g/mol
n(C6H5CL) = m/M = 22.5 / 112.5 = 0.2 mol
n(C6H5CL) = n(C6H5ONa) = 0.2 mol
M(C6H5ONa) =116 g/mol
m(теор)(C6H5ONa) = n*M = 0.2 * 116 = 23.2 g
η(C6H5ONa) = m(прак)/ m(теор) * 100% = 15.08 / 23.2 * 100% = 65%
ответ 65%