Ta có: \(n_{Al}=\dfrac{5}{27}\approx0,19\left(mol\right)\)
a)PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b)Theo PTHH\(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,19=0,285mol\)
\(\Rightarrow V_{H_2}=0,285.22,4=6,384\left(mol\right)\)
c)Theo PTHH:\(n_{HCl}=\dfrac{6}{2}n_{Al}=3.n_{Al}=3.0,19=0,57mol\)
\(\Rightarrow m_{HCl}=0,57.\left(2+35,5\right)=21,375\left(g\right)\)
Đổi : 100ml=100g
\(\Rightarrow C\text{%_{HCl}}=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{21,375}{100}.100\%=27,375\%\)
d)Theo PTHH:\(n_{AlCl_3}=n_{Al}=0,19mol\)
\(\Rightarrow m_{AlCl_3}=0,19.\left(27+3.35,5\right)=25,365\left(g\right)\)
nAl = \(\dfrac{5}{27}\)= 0,185 mol
2Al + 6HCl ->2 AlCl3 + 3H2
0,185->0,555 ->0,185 ->0,2775
b) VH2 = 0,185 . 22,4 = 4,144 (l)
c) bạn cho Vd2 thì ml chỉ tính được CM thôi nhá!
- V = 100 ml = 0,1 (l)
=>CM = \(\dfrac{0,555}{0,1}\) = 5,55 M
d) mAlCl3 = 0,2775 . 133,5 = 37,04 g
