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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, \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
b, \(n_{CuCl_2}=\dfrac{20,25}{135}=0,15\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{CuCl_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
c, \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)
=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)
\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)
\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
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)
\(n_{FeCl_3}=0.2\cdot0.4=0.08\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(0.08...........0.24..............0.08\)
\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)
\(0.08...........0.04\)
\(m_{Fe_2O_3}=0.04\cdot160=6.4\left(g\right)\)
\(V_{dd_{NaOH}}=\dfrac{0.24}{0.5}=0.48\left(l\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\) (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) : \(n_{NaOH}=3n_{FeCl_3}=0,08.3=0,24\left(mol\right)\)
=> \(V_{NaOH}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)
\(n_{CuSO_4}=0,1.0,625=0,0625mol\\ a)Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
\(b)n_{Cu}=n_{CuSO_4}=0,0625mol\\ m_{Cu}=0,0625.64=4g\\ c)ZnSO_4+2NaOH\rightarrow Zn\left(OH\right)_2+Na_2SO_4\\ Zn\left(OH\right)_2\xrightarrow[]{t^0}ZnO+H_2O\\ n_{ZnO}=n_{Zn\left(OH\right)_2}=n_{ZnSO_4}=0,0625mol\\ m=m_{ZnO}=0,0625.81=5,0625g\)