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a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)
Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
\(n_{NaOH}=\dfrac{12}{40}=0.3\left(mol\right)\)
\(n_{HCl}=\dfrac{7.3}{36.5}=0.2\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có :
\(n_{NaOH}>n_{HCl}\Rightarrow NaOHdư\)
\(n_{NaOH\left(pư\right)}=n_{HCl}=n_{NaCl}=0.2\left(mol\right)\)
\(n_{NaOH\left(dư\right)}=0.3-0.2=0.1\left(mol\right)\)
\(m_{cr}=m_{NaOH\left(dư\right)}+m_{NaCl}=0.1\cdot40+0.2\cdot58.5=15.7\left(g\right)\)
Ta có: \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PTHH: NaOH + HCl ---> NaCl + H2O
Ta thấy: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\)
Vậy NaOH dư, HCl hết.
Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
\(n_{Na_2SO_4}=\dfrac{71.20}{100.142}=0,1\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{100.10,4}{100.208}=0,05\left(mol\right)\)
PTHH: \(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\) => BaCl2 hết, Na2SO4 dư
PTHH: \(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)
0,05<--------0,05---->0,05------->0,1
=> \(\left\{{}\begin{matrix}m_{Na_2SO_4}=\left(0,1-0,05\right).142=7,1\left(g\right)\\m_{NaCl}=0,1.58,5=5,85\left(g\right)\end{matrix}\right.\)
mdd sau pư = 71 + 100 - 0,05.233 = 159,35(g)
=> \(\left\{{}\begin{matrix}C\%\left(Na_2SO_4\right)=\dfrac{7,1}{159,35}.100\%=4,456\%\\C\%\left(NaCl\right)=\dfrac{5,85}{159,35}.100\%=3,67\%\end{matrix}\right.\)
Bài 6:
\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)
PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)
_______0,2________0,6______0,2 (mol)
a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)
b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)
Bài 7:
\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
______0,1______0,1_______0,1 (mol)
a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)
b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)
a) nFe= 16/56 =~ 0,3 mol
mH2S04 =( C% .mdd ) /100%= ( 20.100) /100 = 20g
nH2SO4 = 20/98 =~ 0,2mol
lập pthh của pu
Fe + H2SO4 ----------> FeSO4 + H2
1mol 1mol 1mol 1mol
0,3mol 0,2mol
xét tỉ lệ nFe dư sau pư vậy tính theo mol H2SO4
nFe (pư) = (0,2 .1 )/1 =0,2mol
nFe (dư) = 0,3 -0,2 =0,1mol
mFe dư = 0,1 . 56 = 5,6 g
mFeSO4 = 0,2 .152 = 30,4 g
b) mdd sau pư = mFe + m dung môi = 16 +100=116 g
c% Fe = (5,6 / 116) .100%=~ 4,83%
c% FeSO4 =(30,4/116).100%=~ 26,21%
a) đối 200ml =0,2 lít
CMFe =n/v = 0,1 / 0,2 =0,5 mol/lít
CMFeSO4 =n/v = 0,2/0,2=1 mol /lít
\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
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)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\\n_{H_2SO_4}=\dfrac{784\cdot10\%}{98}=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,6}{1}< \dfrac{0,8}{1}\) \(\Rightarrow\) H2SO4 còn dư, Fe phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=n_{H_2}=0,6mol\\n_{H_2SO_4\left(dư\right)}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,2\cdot98=19,6\left(g\right)\\m_{H_2}=0,6\cdot2=1,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=816,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{91,2}{816,4}\cdot100\%\approx11,17\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{19,6}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)
\(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
\(n_{H2SO4}=\dfrac{784.10\%}{98}=0,8\left(mol\right)\)
PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\uparrow\)
Theo pthh : \(n_{H2}=n_{FeSO4}=n_{H2SO4\left(pứ\right)}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow n_{H2SO4\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)
Áp dụng ĐLBTKL :
mFe + m(dd H2SO4) = m(ddspu) + mH2
=> 33,6 + 784 = m(ddspu) + 0,6.2
=> m(ddspu) = 816,4(g)
\(\Rightarrow\left\{{}\begin{matrix}C\%FeSO_{\text{4}}=\dfrac{0,6.152}{816,4}\cdot100\%\approx11,17\%\\C\%H_2SO_{4\left(dư\right)}=\dfrac{0,2.98}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)
nFe=2,8/56=0,05 mol
Fe +2HCl=>FeCl2 +H2
0,05 mol =>0,05 mol=>0,05 mol
mdd sau pứ=2,8+60-0,1=62,7 gam
mFeCl2=127.0,05=6,35 gam
C% dd FeCl2=6,35/62,7.100%=10,13%