cho 10,2 g Al2O3 tac dung vua voi 200 g dung dich HCl
a,Tinh C% dung dich HCl da dung
b, tinh C% dung dich sau phan ung
(giup mình vs !)
cho 10,2 g Al2O3 tac dung vua voi 200 g dung dich HCl
a,Tinh C% dung dich HCl da dung
b, tinh C% dung dich sau phan ung
(giup mình vs !)
n Al2O3=$\frac{10,2}{102}$=0,1 mol
Al2O3+6HCl→2AlCl3+3H2O
0,1→ 0,6 0,2 mol
a.
m ct HCl=0,6.36,5=21,9 g
C% HCl=$\frac{21,9}{200}$.100=10,95 %
b.
m ct AlCl3=0,2.133,5=26,7 g
mdd sau=m Al2O3+m dd HCl
=10,2+200=210,2 g
C% AlCl3=$\frac{26,7}{210,2}$.100≈12,7 %
———————–Nguyễn Hoạt————————–
$n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol \\PTHH : \\Al_2O_3+6HCl\to 2AlCl_3+3H_2O \\a.Theo\ pt : \\n_{HCl}=6.n_{Al_2O_3}=6.0,1=0,6mol \\⇒m_{HCl}=0,6.36,5=21,9g \\⇒C\%_{HCl}=\dfrac{21,9}{200}.100\%=10,95\% \\b.Theo\ pt : \\n_{AlCl_3}=2.n_{Al_2O_3}=2.0,1=0,2mol \\⇒m_{AlCl_3}=0,2.133,5=26,7g \\m_{dd\ spu}=10,2+200=210,2g \\⇒C\%_{AlCl_3}=\dfrac{26,7}{210,2}.100\%=12,7\%$