\(n_{Fe_3O_4}=\frac{m}{M}=\frac{5,8}{232}=0,025\left(mol\right)\)
\(m_{HCl}=\frac{150.3,65}{100}=5,475\left(g\right)\\ \rightarrow n_{HCl}=\frac{m}{M}=\frac{5,475}{36,5}=0,15\left(mol\right)\)
\(PTHH:Fe_3O_4+6HCl\rightarrow FeCl_2+2FeCl_3+3H_2O\)
(mol) 1 6 1 2 3
(mol) 0,025 0,15 0,025 0,05 0,075
Tỉ lệ: \(\frac{0,025}{1}=\frac{0,15}{6}\)
\(m_A=m_{FeCl_2}+m_{FeCl_3}=0,025.127+0,05.162,5=11,3\left(g\right)\)
\(m_{ddA}=m_{Fe_3O_4}+m_{ddHCl}=5,8+150=155,8\left(g\right)\\ \rightarrow V_{ddA}=\frac{m}{D}=\frac{155,8}{1,1}=141,63\left(ml\right)=0,141\left(l\right)\)
\(C_{MddA}=\frac{n}{V}=\frac{0,025+0,05}{0,141}=0,53\left(M\right)\)
\(C\%_{ddA}=\frac{11,3}{155,8}.100\%=7,25\left(\%\right)\)
Từ chỗ PTHH xuống nhé :)
\(PTHH:Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
(mol) 1 8 1 2 4
(mol) 0,01875 3/1280 3/640
Tỉ lệ: \(\frac{0,025}{1}>\frac{0,15}{8}\rightarrow Fe_3O_4.dư\)
\(m_A=m_{FeCl_2}+m_{FeCl_3}=\frac{3}{1280}.127+\frac{3}{640}.162,5=1,06\left(g\right)\)
\(m_{ddA}=5,8+150=155,8\left(g\right)\Rightarrow V_{ddA}=0,141\left(l\right)\)
\(C\%_{ddA}=\frac{1,06}{155,8}.100\%=0,68\left(\%\right)\)
\(C_M=\frac{\frac{3}{1280}+\frac{3}{640}}{0,141}=0,05\left(M\right)\)
