\(n_{MgCl_2}=\dfrac{9,5}{95}=0,1(mol)\\ a,MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl\\ b,n_{Mg(OH)_2}=0,1(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,1.58=5,8(g)\)
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