nH2=$\frac{1,68}{22,4}$=0,075(mol)
PTHH: 2M+2H2O→2MOH+H2↑
0,15 0,15 0,075
MOH+HCl→MCl+H2O
0,15 0,15 0,15
a) theo pt=>nM=2.nH2=0,15(mol)
=>mM=0,15.M=3.45
=>M=23 (hợp lí)
Vậy M là natri (Na)
b) mdd sau pư=3,45+55,2-0,075.2=58,5(g)
mA=mNaCl=0,15.58,5=8,775(g)
=>C%NaCl=$\frac{8,775}{58,5}$.100%=15%
c) theo pt=>nHCl=nMOH=0,15(mol)
=>mHCl=0,15.36,5=5,475(g)
=>C%HCl=$\frac{5,475}{55,2}$.100%=9,92%