解ï¼
1设æMg x g,å³x/24 mol
Al 10.2-x g,å³(10.2-x)/27 mol
(HCl)=250x10^-3 L X 4mol/L=1 mol
Mg + 2 HCl = MgCl2 + H2 â
1 2
x/24
2 Al + 6 HCl = 2 AlCl3 + 3 H2 â
2 6
(10.2-x)/27
2(x/24)+3(10.2-x)/27=1
x=4.8g
10.2-x=5.4g
æ以10.2gééåéä¸é4.8å
ï¼é5.4å
ã
2
n(MgCl2)=n(Mg)=4.8g/(24g/mol)=0.2 mol
n(AlCl3)=n(Al)=5.4g/(27g/mol)=0.2 mol
MgCl2 + 2 NaOH = Mg(OH)2 â + 2 NaCl
1 2
0.2
AlCl3 + 3 NaOH = Al(OH)3 â + 3 NaCl
1 3
0.2
n(NaOH)=2 n(MgCl2)+3 n(AlCl3)=0.2X2+0.2X3=1mol
V(NaOH)=n(NaOH)/c(NaOH)=1mol/(2mol/L)=0.5L=500ml
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