дано
m(C6H6) = 117 g
m практ (C6H5CL) = 118.125 g
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η(C6H5CL) - ?
C6H6+CL2-->C6C5CL+HCL
M(C6H6) = 78 g/mol
n(C6H6) = m/M = 117 / 78 = 1.5 mol
n(C6H6) =n(C6H5CL) = 1.5 mol
M(C6H5CL) = 112.5 g/mol
m теор (C6H5CL) = n*M = 1.5 * 112.5 = 168.75 g
η(C6H5CL) = 118.125 / 168.75 * 100% = 70%
ответ 70%
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Answers & Comments
дано
m(C6H6) = 117 g
m практ (C6H5CL) = 118.125 g
-----------------------
η(C6H5CL) - ?
C6H6+CL2-->C6C5CL+HCL
M(C6H6) = 78 g/mol
n(C6H6) = m/M = 117 / 78 = 1.5 mol
n(C6H6) =n(C6H5CL) = 1.5 mol
M(C6H5CL) = 112.5 g/mol
m теор (C6H5CL) = n*M = 1.5 * 112.5 = 168.75 g
η(C6H5CL) = 118.125 / 168.75 * 100% = 70%
ответ 70%