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a,Hiện tượng: Màu vàng nâu của dung dịch FeCl3 nhạt dần và xuất hiện kết tủa nâu đỏ Fe(OH)3.
\(m_{FeCl_3}=100.13\%=13\left(g\right)\Rightarrow n_{FeCl_3}=\dfrac{13}{162,5}=0,08\left(mol\right)\)
PTHH: 3NaOH + FeCl3 → 3NaCl + Fe(OH)3
Mol: 0,24 0,08 0,24 0,08
b, \(m=m_{ddNaOH}=\dfrac{0,24.40.100\%}{10\%}=96\left(g\right)\)
mNaCl = 0,24.58,5 = 14,04 (g)
mddNaCl = 96 + 100 - 0,08.107 = 187,44 (g)
\(C\%_{ddNaCl}=\dfrac{14,04.100\%}{187,44}=7,49\%\)
\(m_{FeCl_3}=\frac{200.16,25\%}{100\%}=32,5\left(g\right)\)
\(n_{FeCl_3}=\frac{32,5}{162,5}=0,2\left(mol\right)\)
\(PTHH:FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(\left(mol\right)\)____\(0,2\)______\(0,6\)________\(0,2\)_______\(0,6\)
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) \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
\(m_{ddNaCl}=m_{ddFeCl_3}+m_{ddNaOH}-m_{Fe\left(OH\right)_3}=200+120-21,4=298,6\left(g\right)\)
\(C\%_{NaCl}=\frac{35,1}{298,6}.100\%=11,75\%\)
\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
______0,05------>0,15--------->0,05
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)
\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)
PTHH: Fe2(SO4)3 + 6NaOH --> 2Fe(OH)3\(\downarrow\) + 3Na2SO4
________0,05----------------------->0,1
=> mFe(OH)3 = 0,1.107=10,7(g)
nNa2CO3 = 10,6 / 106 = 0,1 (mol)
Na2CO3 + 2CH3COOH -> 2CH3COONa + H2O + CO2
0,1 0,2 0,2 0,1
mdd CH3COOH = 0,2 * 60 / 5 * 100 = 240 (gam)
CO2 + Ca(OH)2 -> CaCO3 + H2O
0,1 0,1
mCaCO3 = 0,1 * 100 = 10 (gam)
mdd = 240 + 10,6 - 0,1 * 44 = 246,2 (gam)
C% = 82 * 0,2 / 246,2 * 100% = 6,66%
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
Na2CO3 + 2CH3COOH ---> 2CH3COONa + CO2 + H2O
0,1---------->0,2----------------->0,2--------------->0,1
CO2 + Ca(OH)2 ---> CaCO3 + H2O
0,1------------------------->0,1
=> \(\left\{{}\begin{matrix}m_{ddCH_3COOH}=\dfrac{0,2.60}{5\%}=240\left(g\right)\\m_{CaCO_3}=0,1.100=10\left(g\right)\end{matrix}\right.\)
\(m_{dd}=10,6+240-0,1.44=246,2\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{82.0,2}{246,2}.100\%=6,66\%\)
\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)
\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,1\cdot160=16\left(g\right)\\ b,n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{200}\cdot100\%=4,9\%\)
Sau phản ứng, thu được hỗn hợp kim loại, suy ra kẽm dư.
$n_{CuSO_4} = \dfrac{80.30\%}{160} = 0,15(mol)$
$Zn + CuSO_4 \to ZnSO_4 + Cu$
$n_{Zn\ pư} = n_{CuSO_4} = 0,15(mol)$
$\Rightarrow m_{Zn\ pư} = 0,15.65 = 9,75(gam)$
Sau phản ứng, $m_{dd} = 9,75 + 80 - 0,15.64 = 80,15(gam)$
$C\%_{ZnSO_4} = \dfrac{0,15.161}{80,15}.100\% = 30,13\%$
\(a,PTHH:FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe\left(OH\right)_2}=n_{FeCl_3}=\dfrac{10,7}{107}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_3}}=0,1\cdot162,5=16,25\left(g\right)\\ \Rightarrow m_{dd_{FeCl_3}}=\dfrac{16,25\cdot100\%}{5\%}=325\left(g\right)\\ b,n_{NaOH}=n_{NaCl}=3n_{Fe\left(OH\right)_3}=0,3\left(mol\right)\\ \Rightarrow m_{NaOH}=0,3\cdot40=12\left(g\right)\\ m_{NaCl}=0,3\cdot58,5=17,55\left(g\right)\\ \Rightarrow m_{dd_{NaCl}}=325+150-10,7=464,3\left(g\right)\\ \Rightarrow C\%_{dd_{NaCl}}=\dfrac{17,55}{464,3}\cdot100\%\approx3,78\%\)