a) \(n_{H_2SO_4}=0,05.2=0,1\left(mol\right)\)
PTHH: \(2MOH+H_2SO_4\rightarrow M_2SO_4+2H_2O\)
0,2<-------0,1--------->0,1
=> \(M_{MOH}=\dfrac{8}{0,2}=40\left(g/mol\right)\)
=> MM = 40 - 17 = 23 (g/mol)
=> M là Natri (Na)
b) \(C_{M\left(Na_2SO_4\right)}=\dfrac{0,1}{0,05}=2M\)
c) \(m_{dd}=1,04.50=52\left(g\right)\)
=> \(C\%_{Na_2SO_4}=\dfrac{0,1.142}{52}.100\%=27,31\%\)