2Na+2H2O\(\rightarrow\)2NaOH+H2
Ba+2H2O\(\rightarrow\)Ba(OH)2+H2
NaOH+HCl\(\rightarrow\)NaCl+H2O
Ba(OH)2+2HCl\(\rightarrow\)BaCl2+2H2O
- Gọi x,y là số mol Na và Ba. Theo các PTHH ta có hệ:
23x+137y=0,594
58,5x+208y=0,949
Giải hệ ta có: x=0,002, y=0,004
\(n_{H_2}=0,5x+y=\)0,005mol
\(V_{H_2}=0,005.22,4=0,112l\)
\(n_{HCl}=x+2y=0,01mol\)
\(C_{M_{HCl}}=\dfrac{0,01}{0,1}=0,1M\)
mNa=0,002.23=0,046g
mBa=0,004.137=0,548g
2Na + 2H2O \(\rightarrow\)2NaOH + H2 (1)
Ba + 2H2O \(\rightarrow\)Ba(OH)2 + H2 (2)
NaOH + HCl \(\rightarrow\)NaCl + H2O (3)
Ba(OH)2 + 2HCl \(\rightarrow\)BaCl2 + 2H2O (4)
Đặt nNa=a
nBa=b
Ta có:
\(\left\{{}\begin{matrix}23a+137b=0,594\\58,5a+208b=0,949\end{matrix}\right.\)
=>a=0,002;b=0,004
Theo PTHH 1 và 2 ta có:
\(\dfrac{1}{2}\)nNa=nH2=0,001(mol)
nBa=nH2=0,004(mol)
VH2=0,005.22,4=0,112(lít)
mNa=23.0,002=0,046(g)
mBa=0,594-0,046=0,548(g)
mCl=0,949-0,594=0,355(g)
nCl=\(\dfrac{0,355}{35,5}=0,01\left(mol\right)\)
Ta cso:
nCl=nHCl=0,01(mol)
CM dd HCl=\(\dfrac{0,01}{0,1}=0,1M\)
