Bài 4:
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{HCl}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\C\%_{HCl}=\dfrac{0,3\cdot36,5}{100}\cdot100\%=10,95\%\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
nAl = 2,7 /27=0,1 mol
2Al + 6HCl --> 2AlCl3 + 3H2
0,1 0,3 0,1 0,15 mol
=>mHCl = 0,3*36,5=10,95 g
C% HCl=10,95*100/100=10,95%
=> mAlCl3 = 0,1 * 133,5=13,35 g
mH2=0,15*2=0,3 g
