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Bài 10:
PTHH: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\downarrow\)
a) Ta có: \(n_{Na_2CO_3}=\dfrac{200\cdot10,6\%}{106}=0,2\left(mol\right)=n_{BaCO_3}\)
\(\Rightarrow m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\)
b) Theo PTHH: \(n_{BaCl_2}=n_{BaCO_3}=0,2mol\)
\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2\cdot208}{120}\cdot100\%\approx34,67\%\)
c) Theo PTHH: \(n_{NaCl}=2n_{BaCl_2}=0,4mol\) \(\Rightarrow m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
Mặt khác: \(m_{dd}=m_{ddNa_2CO_3}+m_{ddBaCl_2}-m_{BaCO_3}=280,6\left(g\right)\)
\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{280,6}\cdot100\%\approx8,34\%\)
\(a,PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \left\{{}\begin{matrix}m_{H_2SO_4}=\dfrac{300\cdot9,8\%}{100\%}=29,4\left(g\right)\\m_{BaCl_2}=\dfrac{200\cdot26\%}{100\%}=52\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\end{matrix}\right.\)
Vì \(\dfrac{n_{H_2SO_4}}{1}>\dfrac{n_{BaCl_2}}{1}\) nên H2SO4 dư
\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow a=m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\ b,n_{HCl}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,25\cdot36,5=9,125\left(g\right)\\ m_{dd_{HCl}}=300+200-58,25=441,75\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{9,125}{441,75}\cdot100\%\approx2,07\%\)
a) Vì: mA < 400 (g) nên phải có khí thoát ra → muối có dạng MHSO4 và khí là: CO2
b)
c) Tác dụng được với: MgCO3, Ba(HSO3)2, Al2O3, Fe(OH)2, Fe, Fe(NO3)2
Pt: 2NaHSO4 + MgCO3 → Na2SO4 + MgSO4 + CO2↑ + H2O
2NaHSO4 + Ba(HSO3)2 → BaSO4 + Na2SO4 + SO2↑ + 2H2O
6NaHSO4 + Al2O3 → 3Na2SO4 + Al2(SO4)3 + 3H2O
2NaHSO4 + Fe(OH)2 → Na2SO4 + FeSO4 + 2H2O
2NaHSO4 + Fe → Na2SO4 + FeSO4 + H2↑
12NaHSO4 + 9Fe(NO3)2 → 5Fe(NO3)3 + 2Fe2(SO4)3 + 6Na2SO4 + 3NO↑ + 6H2O
\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)
\(a,\left\{{}\begin{matrix}m_{BaCl_2}=\dfrac{100\cdot10,4\%}{100\%}=10,4\left(g\right)\\m_{H_2SO_4}=\dfrac{200\cdot9,8\%}{100\%}=19,6\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\\n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\end{matrix}\right.\)
\(PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
Vì \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{2}\) nên \(H_2SO_4\) dư
\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{BaSO_4}=0,05\cdot233=11,65\left(g\right)\)
\(b,n_{HCl}=n_{BaSO_4}=0,05\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,05\cdot36,5=1,825\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=100+200-11,65=288,35\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{1,825}{288,35}\cdot100\%\approx0,63\%\)
\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)
\(n_{Na_2SO_4}=\dfrac{100.28,4\%}{142}=0,2\left(mol\right)\)
PTHH: \(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)
0,2------>0,2--------->0,2
`=>` \(\left\{{}\begin{matrix}C\%_{BaCl_2}=\dfrac{0,2.208}{200}.100\%=20,8\%\\m=m_{BaSO_4}=0,2.233=46,6\left(g\right)\end{matrix}\right.\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
a. PTHH: Fe3O4 + 4H2SO4 ---> FeSO4 + Fe2(SO4)3 + 4H2O
Theo PT: \(n_{H_2SO_4}=4.n_{Fe_3O_4}=4.0,01=0,04\left(mol\right)\)
=> \(m_{H_2SO_4}=0,04.98=3,92\left(g\right)\)
Theo đề, ta có: \(C_{\%_{H_2SO_4}}=\dfrac{3,92}{m_{dd_{H_2SO_4}}}.100\%=20\%\)
=> \(m_{dd_{H_2SO_4}}=19,6\left(g\right)\)
b. Ta có: \(m_{dd_{SauPỨ}}=2,32+19,6=21,92\left(g\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,01\left(mol\right)\)
=> \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,01.400=4\left(g\right)\)
=> \(m_{SauPỨ}=1,52+4=5,52\left(g\right)\)
=> \(C_{\%_{SauPỨ}}=\dfrac{5,52}{21,92}.100\%=25,18\%\)
\(n_{H_2SO_4}=\dfrac{200.4,9\%}{98}=0,1\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ a.n_{BaSO_4}=n_{BaCl_2}=n_{H_2SO_4}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ C\%_{ddBaCl_2}=\dfrac{0,1.208}{300}.100\approx6,933\%\\ b.m_{ddsau}=200+300-0,1.233=476,7\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{476,7}.100\approx1,531\%\)