Ta có: \(C\%=\frac{m_{ct}}{m_{dd}}.100\%\)
\(\Rightarrow m_{ct}=\frac{m_{dd}.C\%}{100\%}\)
\(\Rightarrow m_{HCl}=\frac{10\%.200}{100\%}=20\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{20}{36,5}=\frac{40}{73}\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PTHH: \(n_{HCl}:n_{AlCl_3}=6:2=3:1\)
\(\Rightarrow n_{AlCl_3}=n_{HCl}.\frac{1}{3}=\frac{40}{73}.\frac{1}{3}=0,18\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,18.133,5=24,03\left(g\right)\)
b) Theo PTHH: \(n_{HCl}:n_{H_2}=6:3=2:1\)
\(\Rightarrow n_{H_2}=n_{HCl}.\frac{1}{2}=\frac{40}{73}.\frac{1}{2}=0,27\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,27.22,4=6,048\left(l\right)\)
mHCl = 20 g
nHCl = 40/73 (mol)
2Al + 6HCl --> 2AlCl3 + 3H2
Từ PTHH :
=> nAlCl3 = 40/219 mol => mAlCl3 = 24.38 g
=> nH2 = 20/73 mol => VH2 = 6.14 l