a) nNaOH = 0,1.1 = 0,1 (mol)
PTHH: \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
0,05<------0,1----------->0,05----->0,1
=> m = 0,05.58 = 2,9 (g)
b) \(V=\dfrac{0,05}{2}=0,025\left(l\right)\)
\(C_{M\left(NaCl\right)}=\dfrac{0,1}{0,025+0,1}=0,8M\)