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DmitryImmortal Решение задачи см. во вложении

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2C6H5OH+2Na->2C6H5ONa+H2
n=m\M
M(C6H5OH)=94 г\моль
n(C6H5OH)=9,4\94=0,1 моль
0,5*n(C6H5OH)=n(H2)=0,5*0,1=0,05 моль
v=n*Vm
Vm=22,4 дм3\моль
V(H2)=0,05*22,4=1,12 дм3 (л)

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