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\(n_{FeCl_3}=0.2\cdot0.4=0.08\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(0.08...........0.24..............0.08\)
\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)
\(0.08...........0.04\)
\(m_{Fe_2O_3}=0.04\cdot160=6.4\left(g\right)\)
\(V_{dd_{NaOH}}=\dfrac{0.24}{0.5}=0.48\left(l\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\) (1)
\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\) (2)
\(n_{FeCl_3}=0,2.0,4=0,08\left(mol\right)\)
Bảo toàn nguyên tố Fe : \(n_{FeCl_3}=2n_{Fe_2O_3}=0,08\left(mol\right)\)
=> \(n_{Fe_2O_3}=0,04\left(mol\right)\)
=> \(m_{Fe_2O_3}=0,04.160=6,4\left(g\right)\)
Theo PT (1) : \(n_{NaOH}=3n_{FeCl_3}=0,08.3=0,24\left(mol\right)\)
=> \(V_{NaOH}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)
\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)
=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)
\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)
\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)
=> m Fe2O3 = 0,1 . 160=16(g)
a)
\(FeSO_4 + 2NaOH \to Fe(OH)_2 + Na_2SO_4\)
b)
\(n_{FeSO_4} = 0,4.0,5 = 0,2(mol) ; n_{NaOH} = 0,5.0,5 = 0,25(mol)\)
Ta thấy : \(2n_{FeSO_4} = 0,4 > n_{NaOH} = 0,25\) nên FeSO4 dư.
Theo PTHH :
\(n_{Fe(OH)_2} = 0,5n_{NaOH} = 0,125(mol)\\ \Rightarrow m_{Fe(OH)_2} = 0,125.90 = 11,25(gam)\)
c)
\(4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O\)
Theo PTHH :
\(n_{Fe_2O_3} = 0,5n_{Fe(OH)_2} = 0,0625(mol)\\ \Rightarrow m_{Fe_2O_3} = 0,0625.160 = 10(gam)\)
a) PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,5\cdot0,1=0,05\left(mol\right)\\n_{FeCl_3}=0,2\cdot0,2=0,04\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,05}{3}< \dfrac{0,04}{1}\) \(\Rightarrow\) NaOH p/ứ hết, FeCl3 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,05\left(mol\right)\\n_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,05\cdot58,5=2,925\left(g\right)\\m_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\cdot162,5\approx3,8\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)
Ta có: \(n_{Fe\left(OH\right)_3}=\dfrac{1}{60}\left(mol\right)\) \(\Rightarrow n_{H_2O}=\dfrac{1}{40}\left(mol\right)\) \(\Rightarrow m_{H_2O}=\dfrac{1}{40}\cdot18=0,45\left(g\right)\)