Ba(OH)2+MgSO4->BaSO4+Mg(Oh)2
0,2-------------------------0,2-------0,2
nBa(oh)2=0,2 mol
=>mkt=0,2.233+0,2.58=58,2g
\(n_{Ba\left(OH\right)_2}=0,2.1=0,2\left(mol\right)\)
\(PTHH:Ba\left(OH\right)_2+MgSO_4\rightarrow Mg\left(OH\right)_2\downarrow+BaSO_4\downarrow\)
`(mol)`______`0,2`_____`0,2`_________`0,2`_______`0,2`____
\(m_{\downarrow}=m_{Mg\left(OH\right)_2}+m_{BaSO_4}=0,2\left(58+233\right)=58,2\left(g\right)\)