дано

m(C2H5OH) = 4.6 g

m(ppa Ba(OH)2) = 540 g

W(Ba(OH)2) = 10%

---------------------------

m((C2H5O)2Ba)-?

m(Ba(OH)2) = 540*10% / 100% = 54 g

M(C2H5OH) = 46 g/mol

n(C2H5OH)  = m/M = 4.6 / 46 = 0.1 mol

M(Ba(OH)2) = 171 g/mol

n(Ba(OH)2) = m/M = 54 / 171 = 0.32 mol

n(C2H5OH) < n(Ba(OH)2)

2C2H5OH+Ba(OH)2-->(C2H5O)2Ba+2H2O

2n(C2H5OH) = n((C2H5O)2Ba)

n((C2H5O)2Ba) = 0.1 / 2 = 0.05 mol

M((C2H5O)2Ba) = 227 g/mol

m((C2H5O)2Ba)  = n*M = 0.05 * 227 = 11.35 g

ответ 11.35 г