a)\(n_{Al_2O_3}=\frac{m}{M}=\frac{21,42}{102}=0,21mol\)
PTHH:
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,21 0,36 0,31 0,36(mol)
b)\(m_{Al_2\left(SO_4\right)_3}=n.M=0,21.342=71,82g\)
c)\(V_{ddH_2SO_4}=\frac{n}{C_M}=\frac{0,36}{2}=0,18l=180ml\)
\(m_{ddH_2SO_4}=180.1,14=205,2g\)
a, Al2O3 + 3H2SO4 => Al2(SO4)3 + 3H2O
0,21 => 0,63 0,21
b, mol al203 =21,42/102 =0,21 mol
m al2(so4)3 =0,21x 342 =71,82g
c, Cm h2so4 =n/v => 2 =0,63/v => V h2s04 =0,63/2 =0,315lít =315 ml
ta có d =m/v => m = d x V=1,14x 315=359,1 g
a,PTHH : Al2O3+3H2SO4--->Al2(SO4)3+3H2O
b, Ta có : \(n_{Al2O3}=\frac{21,42}{102}=0,21\left(mol\right)\)
\(\Rightarrow m_{Al2O3}=21,42.342=21,82\)
c,Vdd = 180l
=>mdd = 180.1,14= 205,2(g)
nAl2O3 = 0,21 mol
Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
0,21 0,63 0,21 (mol)
mAl2(SO4)3 = 0,21.342 = 71,82 g
VddH2SO4 = 0,63/2 = 0,315 l = 315 ml
mH2SO4 = 315.1,14 = 359,1 g
