a) \(n_{MgCl_2}=\dfrac{9,5}{95}=0,1\left(mol\right)\); \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
PTHH: \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) => MgCl2 hết, NaOH dư
PTHH: \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
0,1------->0,2---------->0,1-------->0,2
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
0,1----->0,1
=> mMgO = 0,1.40 = 4 (g)
b) \(\left\{{}\begin{matrix}m_{NaCl}=0,2.58,5=11,7\left(g\right)\\m_{NaOH\left(dư\right)}=\left(0,4-0,2\right).40=8\left(g\right)\end{matrix}\right.\)