Đáp án:
Giải thích các bước giải:
4/ `m_{dd}=300+200=500\ gam`
`n_{HCl}=\frac{300.7,3\%}{36,5}=0,6\ mol`
`n_{Ba(OH)_2}=\frac{200.17,1\%}{171}=0,2\ mol`
`Ba(OH)_2+2HCl\to BaCl_2+2H_2O`
`\Rightarrow n_{HCl\text{dư}}= n_{BaCl_2}=0,2\ mol`
`C\%_{HCl}=\frac{0,2.36,5.100}{500}=1,46\%`
`C\%_{BaCl_2}=\frac{0,2.208.100}{500}=8,32\%`
5/ dd A: `AlCl_3, FeCl_2, $HCl`
rắn B: `Cu`
dd C: `NaAlO_2, NaCl, NaOH`
rắn D: `Fe(OH)_2`
rắn E: `Fe_2O_3`
PTHH:
1/ $2Al+6HCl\to 2AlCl_3+3H_2$
2/ $Fe+2HCl\to FeCl_2+H_2$
3/ $NaOH+HCl\to NaCl+H_2O$
4/ $AlCl_3+3NaOH\to Al(OH)_3+3NaCl$
5/ $Al(OH)_3+NaOH\to NaAlO_2+2H_2O$
6/ $FeCl_2+2NaOH\to Fe(OH)_2+2NaCl$
7/ $2Fe(OH)_2+\dfrac{1}{2}O_2\buildrel{{t^o}}\over\to Fe_2O_3+2H_2O$
