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\(n_{BaCl_2}=\frac{400.5,2\%}{208}=0,1\left(mol\right);n_{H_2SO_4}=\frac{100.1,14.19,6\%}{98}=0,228\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
Theo đề: 0,1.........0,228.....................................(mol)
Lập tỉ lệ: \(\frac{0,1}{1}< \frac{0,228}{1}\)=> Sau phản ứng H2SO4 dư
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
dd sau khi lọc bỏ kết tủa: H2SO4 dư, HCl
\(m_{ddsaup.ứ}=400+114-23,3=490,7\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\frac{\left(0,228-0,1\right).98}{490,7}.100=2,56\%\)
\(C\%_{HCl}=\frac{0,1.2.36,5}{490,7}.100=1,49\%\)
\(n_{H_2SO_4}=0.1\cdot2=0.2\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=100\cdot1.2=120\left(g\right)\)
\(n_{BaCl_2}=0.1\cdot1=0.1\left(mol\right)\)
\(m_{dd_{BaCl_2}}=100\cdot1.32=132\left(g\right)\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
\(0.1................0.1.........0.1...............0.2\)
\(\Rightarrow H_2SO_4dư\)
\(m_{BaSO_4}=0.1\cdot233=23.3\left(g\right)\)
\(V_{dd}=0.1+0.1=0.2\left(l\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2-0.1}{0.2}=0.5\left(M\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)
\(m_{\text{dung dịch sau phản ứng}}=120+132-23.3=228.7\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0.1\cdot98}{228.7}\cdot100\%=4.28\%\)
\(C\%_{HCl}=\dfrac{0.2\cdot36.5}{228.7}\cdot100\%=3.2\%\)
a) nH2SO4 = 0,2 mol
nBaCl2 = 0,1 mol
H2SO4 (0,1) + BaCl2 (0,1) -----> BaSO4 (0,1) + 2HCl (0,2)
- Theo PTHH: nBaSO4 = 0,1 mol
=> mBaSO4 = 23,3 gam
b) - dd sau phản ứng gồm: \(\left\{{}\begin{matrix}HCl:0,2\left(mol\right)\\H2SO4_{dư}:0,1\left(mol\right)\end{matrix}\right.\)
mdd sau = \(100.1,2+100.1,32-23,3=228,7\left(gam\right)\)
=> \(C\%HCl=\dfrac{0,2.36,5.100}{228,7}=3,192\%\)
=> \(C\%H2SO4_{dư}=\dfrac{0,1.98.100}{228,7}=4,2851\%\)
V dd sau = 0,1 + 0,1 = 0,2 lít
=> CM HCl = 0,2/0,2 = 1M
=> CM H2SO4 dư = 0,1 / 0,2 = 0,5M
fe + cuso4 ---> cu + feso4
nfe=0,035, CMcuso4=(10*10*1.12)/160=0,7, ncuso4=0,07
nfe=0,035 < ncuso4=0,07 ===> cuso4 dư
dd gồm có feso4, cuso4 dư
CMcuso4dư=(0,07-0,035)/0.1=0.35M
CMfeso4=0,035/0,1=0,35M
\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)
\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)
\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)
Đã sửa lần cuối lúc 20:45
5. \(n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\)
\(m_{dd}=1,14.100=114\left(g\right)\)
\(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{22,8}{98}=0,23\left(mol\right)\)
\(Pt:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
0,1 mol 0,23mol \(\rightarrow0,1mol\) \(\rightarrow0,2mol\)
Lập tỉ số: \(n_{BaCl_2}\) : \(n_{H_2SO_4}=0,1< 0,23\)
\(\Rightarrow BaCl_2\) hết; \(H_2SO_4\) dư
\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
\(n_{H_2SO_4\left(dư\right)}=0,23-0,1=0,13\left(mol\right)\)
\(\Sigma_{hh\left(spu\right)}=400+114-23,3=490,7\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,13.98.100}{490,7}=2,6\%\)
\(C\%_{HCl}=\dfrac{0,2.36,5.100}{490,7}=1,49\%\)
6. \(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)
0,5mol 0,4mol
Lập tỉ số: \(n_{NaOH}\) : \(n_{HCl}=0,5>0,4\)
\(\Rightarrow NaOH\) dư; HCl hết
\(n_{NaOH\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(m_{NaOH\left(dư\right)}=0,1.40=4\left(g\right)\)
7. \(n_{Ba\left(OH\right)_2}=\dfrac{300.30\%}{171}=0,52\left(mol\right)\)
Pt: \(Ba\left(OH\right)_2+FeSO_4\rightarrow BaSO_4+Fe\left(OH\right)_2\)
0,52mol \(\rightarrow0,52mol\)\(\rightarrow0,52mol\)
\(m_{BaSO_4}=0,52.233=121,16\left(g\right)\)
\(m_{Fe\left(OH\right)_2}=0,52.90=46,8\left(g\right)\)
\(\Sigma_{hh\left(spu\right)}=300+800-121,16=978,84\left(g\right)\)
\(C\%_{Fe\left(OH\right)_2}=\dfrac{46,8.100}{978,84}=4,78\%\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\)
a+b) Ta có: \(n_{BaCl_2}=\dfrac{400\cdot5,2\%}{208}=0,1\left(mol\right)=n_{H_2SO_4}=n_{BaSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,1\cdot98}{20\%}=49\left(g\right)\\m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{HCl}=0,2\left(mol\right)\) \(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddBaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=425,7\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{425,7}\cdot100\%\approx1,71\%\)
Bạn xem lại giúp mình , coi đề có bị thiếu gì không nhé
\(m_{dd.BaCl_2}=400.1,003=401,2\left(g\right)\)
=> \(n_{BaCl_2}=\dfrac{401,2.5,2\%}{208}=0,1003\left(mol\right)\)
\(m_{dd.H_2SO_4}=100.1,14=114\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{114.20\%}{98}=\dfrac{57}{245}\left(mol\right)\)
PTHH: BaCl2 + H2SO4 --> BaSO4 + 2HCl
Xét tỉ lệ: \(\dfrac{0,1003}{1}< \dfrac{\dfrac{57}{245}}{1}\) => BaCl2 hết, H2SO4 dư
PTHH: BaCl2 + H2SO4 --> BaSO4 + 2HCl
0,1003->0,1003-->0,1003-->0,2006
mdd sau pư = 401,2 + 114 - 0,1003.233 = 491,8301 (g)
\(\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{98\left(\dfrac{57}{245}-0,1003\right)}{491,8301}.100\%=2,637\%\\C\%_{HCl}=\dfrac{0,2006.36,5}{491,8301}.100\%=1,489\%\end{matrix}\right.\)
mddBaCl2 = 1,003 . 400 = 401,2 (g) mBaCl2 = 401,2 . 5,2% = 20,8624 (g)
nBaCl2 = 20,8624/208 = 0,1003 (mol)
mddH2SO4 = 1,14.100 = 114 (g) mH2SO4 = 114 . 20% = 22,8 (g)
nH2SO4 = 22,8/98 (mol)
PTHH: BaCl2 + H2SO4 -> BaSO4 + 2HCl
Bđ: 0,1003 22,8/98
Pư: 0,1003 -> 0,1003 -> 0,1003 -> 0,2006 (mol)
Sau: 0 0,132 0,1003 0,2006 (mol)
Dung dịch sau phản ứng chứa:
mH2SO4 dư = 22,8 - 0,1003.98 = 12,9706 (g)
mHCl = 0,2006.36,5 = 7,3219 (g)
Khối lượng dd sau pư: mdd sau pư = mddBaCl2 + mddH2SO4 - mBaSO4
= 401,2 + 114 - 0,1003.233 = 491,8301 (g)
Nồng độ phần trăm:
C% H2SO4 = (12,9706/491,8301).100% ≈ 2,64%
C% HCl = (7,3219/491,8301).100% ≈ 1,49%