a)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2<---0,6<-------------0,3
=> mAl = 0,2.27 = 5,4 (g)
=> \(m_{Al_2O_3}=15,6-5,4=10,2\left(g\right)\)
b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
0,1----->0,6
=> mHCl = (0,6 + 0,6).36,5 = 43,8 (g)
=> \(m_{dd.HCl}=\dfrac{43,8.100}{20}=219\left(g\right)\)
a)
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
$n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$
$m_{Al_2O_3} = 15,6 - 5,4 = 10,2(gam)$
b)
$n_{Al_2O_3} = 0,1(mol)$
Theo PTHH : $n_{HCl} = 3n_{Al} + 6n_{Al_2O_3} = 1,2(mol)$
$m_{dd\ HCl} = \dfrac{1,2.36,5}{20\%} = 219(gam)$
a)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2<---0,6<------------------0,3
=> \(m_{Al}=0,2.27=5,4\left(g\right)\)
=> \(m_{Al_2O_3}=15,6-5,4=10,2\left(g\right)\)
b)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
0,1------->0,6
=> \(m_{HCl}=\left(0,6+0,6\right).36,5=43,8\left(g\right)\)
=> \(m_{dd.HCl}=\dfrac{43,8.100}{20}=219\left(g\right)\)
