Làm:
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Fe+ 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
P/ư : 0,3 ------ 0,6---------0,3--------0,3 (mol)
\(\Rightarrow m_{Fe}=n.M=0,3.56=16,8\left(g\right)\)
\(\Rightarrow m_{Al_{2_{ }}O_3}=m_{hh}-m_{Fe}=37,2-16,8=20,4\left(g\right)\)
\(\Rightarrow n_{Al_2O_3}=\frac{m}{M}=0,2\left(mol\right)\)
Al2O3 + 6HCl \(\rightarrow\) 2AlCl3 + 3H2O
P/ư: 0,2----------1,2----------0,4---------0,6 (mol)
a,\(m_{hhHCl}=n.M=\left(1,2+0,6\right).36,5=65,7\left(g\right)\)
b,\(\%m_{Fe}=\frac{m_{Fe}.100}{m_{hh}}=45,1\%\)
\(\Rightarrow\)%mAl2O3 = 54,9%
c, \(m_{\text{dd}saupu}=m_{hh}+m_{\text{dd}HCl}-m_{H2}\)
\(=37,2+\frac{65,7.100}{20}-0,3.2\)
\(=365,1\left(g\right)\)
\(C_M_{FeCl2}=\frac{m.100}{m\text{dd}}=34,8\%\)
\(C_MAlCl3=36,6\%\)
Mình làm lại mỗi ý c thôi nha
Làm:
c, \(m_{\text{dd}saupu=}m_{hh}+m_{\text{dd}HCl}-m_{H2}\)
\(=37,2+\frac{65,7.100}{20}-0,3.2\)
\(=365,1\left(g\right)\)
\(C\%_{FeCl2}=\frac{m.100}{m_{\text{dd}}}=34,8\left(g\right)\)
\(C\%_{AlCl3}=36,6\%\)