\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{H2}=0,15\left(mol\right)\)
\(\Rightarrow n_{Al}=0,1\left(mol\right)\)
\(n_{Al2\left(SO4\right)3}=0,05\left(mol\right)\)
\(n_{H2SO4}=0,15\left(mol\right)\)
a,
\(m_{Al}=2,7\left(g\right)\)
\(m_{H2SO4}=14,7\left(g\right)\)
b, \(m_{Al2\left(SO4\right)3}=17,1\left(g\right)\)
a) 2Al+3H2SO4--->Al2(SO4)3+3H2
n H2=3,36/22,4=0,15(mol)
n Al=2/3 n H2=0,1(mol)
m Al=0,1.27=2,7(g)
n H2SO4=n H2=0,15(mol)
m H2SO4=0,15.98=14,7(g)
n Al2(SO4)3=1/3n H2=0,05(mol)
m Al2(SO4)3=0,05.342=17,1(g)
nH2 = VH2 : 22,4 = 3,36 : 22,4 = 0,15 mol
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Tỉ lệ: 2 3
Pứ: 0,1 mol -0,3------0,1--------- 0,15
Từ pthh ta có nAl = 2/3 nH2 = 2/3 . 0,15 = 0,1 mol
=> mAl = nAl . MAl = 0,1 . 27 = 2,7g
=>mHCl=0,3.36,5=10,95 g
=>mAlCl3=0,1.133,5=13,35g
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+4H_2\)
Ta có: \(VH_2=3,36l\Rightarrow n_{H_2}=\frac{3,36}{22,4}=0,15mol\)
theo PTHH:\(n_{Al}=\frac{1}{2}n_{H_2}=0,075mol\)
\(\Rightarrow m_{Al}=2,025g\)
theo PTHH:\(n_{H_2SO_4}=\frac{3}{4}n_{H_2}=0,1125mol\)
\(\Rightarrow m_{H_2SO_4}=11,025g\) b,theo PTHH : \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{4}n_{H_2}=0,375mol\) \(\Rightarrow m_{Al_2\left(SO_4\right)_3}=128,25g\)
