Câu 9:
a,
Gọi chung 2 oxit kim loại là $R_2O$
$R_2O+2HCl\to 2RCl+H_2O$
$n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6(mol)$
$\Rightarrow n_{R_2O}=0,6(mol)$
$\overline{M}=\dfrac{24,36}{0,6}=40,6=2R+16$
$\Leftrightarrow R=12,3$
$\Rightarrow$ 2 kim loại là $Li$, $Na$
b,
Gọi x, y là số mol $Li_2O$, $Na_2O$
$\Rightarrow 30x+62y=24,36$ (1)
Theo PTHH: $x+y=0,6$ (2)
$(1)(2)\Rightarrow x=0,40125; y=0,19875$
$n_{LiCl}=2x=0,8025(mol)\Rightarrow m_{LiCl}=0,8025.42,5=34,10625g$
$n_{NaCl}=2x=0,3975(mol)\Rightarrow m_{NaCl}=0,3975.58,5=23,25375g$
Câu 10:
$n_{H_3PO_4}=\dfrac{137,2.10\%}{98}=0,14(mol)$
$H_3PO_4+3NaOH\to Na_3PO_4+3H_2O$
$\Rightarrow n_{Na_3PO_4}=0,14(mol); n_{NaOH}=0,14.3=0,42(mol)$
$m_{dd NaOH}=0,42.40:20\%=84g$
$\Rightarrow m_{dd\text{spu}}=84+137,2=221,2g$
$\to C\%_{Na_3PO_4}=\dfrac{0,42.164.100}{221,2}=31,1\%$
