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\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)
=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)
\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)
\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
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, 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: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
a) Na2O +H2O-->2NaOH (1)
2NaOH +CuSO4 -->Na2SO4+ Cu(OH)2 (2)
Cu(OH)2 -to-> CuO +H2O (3)
b) mNa2O=8.100/100=8(g)
=>nNa2O=8/62=0,13(mol)
theo(2) :nCu(OH)2=1/2nNaOH=0,065(mol)
theo(3):nCuO=nCu(OH)2=0,065(mol)
=>mCuO=0,065.80=5,2(g)
c) CuO +2HCl-->CuCl2+H2O (4)
theo (4) : nHCl=2nCuO=0,13(mol)
mddHCl 25%=0,13.36,5.100250,13.36,5.10025=18,98(g)
a,\(\text{CuO+H2SO4}\rightarrow\text{CuSO4+H2O}\)
\(\text{CuSO4+2NaOH}\rightarrow\text{Na2SO4+Cu(OH)2}\)
\(\text{Cu(OH)2}\rightarrow\text{CuO+H2O}\)
\(\text{b) nH2SO4=0,02(mol)}\)
mdd spu=0,02x80+20=21,6(g)
\(\text{C%=0,02x160/21,6=14,81%}\)
c) nNaOH=0,04(mol)
mdd=0,04x40/8%=20(g)
\(\text{d) mcr=0,02x80=1,6(g)}\)