MO + 2HCL $\rightarrow$ $MCL_{2}$ +$H_{2}$ O
$R_{2}$ $O_{3}$ + 6HCL $\rightarrow$ 2R$CL_{3}$ +3$H_{2}$ O
2R + 6HCL$\rightarrow$2RC$L_{5}$ +3$H_{2}$
M + 2HCL $\rightarrow$ M$CL_{2}$ +$H_{2}$
n$H_{2}$ =$\dfrac{2,24}{22,4}$ =0,1(md)$\rightarrow$m$H_{2}$ =0,1.2=0,2(9)
nHCL=0,5.1=0,5 (md) $\rightarrow$ nCL = 0,5 (md)
$\rightarrow$mHCL=0,5.36,5=18,25 (g)
Theo pt pứ:
nHCL pứ KL=2n$H_{2}$ = 0,2 (md)
$\rightarrow$ nHCL pứ oxit= 0,5 - 0,2 = 0,3 (md)
$\rightarrow$nHCL pứ oxit = 2n$H_{2}$O$\rightarrow$n$H_{2}$ O= 0,15 (md)
$\rightarrow$n$H_{2}$ ) = 0,15 . 18 = 2,7 (g)
Theo đ/l bảo toàn khối lượng
$m_{muối}$ + m $H_{2}$ O +m$H_{2}$ =m$h^{2}$ +m oxit = 10 + 18,25
$\rightarrow$ $m_{muối}$ = 28,25 - 2,7 - 0,2 = 25,45