\(m_{FeCl_3}=\frac{200.16,25\%}{100\%}=32,5\left(g\right)\)
\(n_{FeCl_3}=\frac{32,5}{162,5}=0,2\left(mol\right)\)
\(PTHH:FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(\left(mol\right)\)____\(0,2\)______\(0,6\)________\(0,2\)_______\(0,6\)
a) \(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\frac{24.100\%}{20\%}=120\left(g\right)\)
b) \(m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)
c) \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
\(m_{ddNaCl}=m_{ddFeCl_3}+m_{ddNaOH}-m_{Fe\left(OH\right)_3}=200+120-21,4=298,6\left(g\right)\)
\(C\%_{NaCl}=\frac{35,1}{298,6}.100\%=11,75\%\)
Fe(OH)3 sao khối lượng mol 107 được em?
