cho 27,4 g Bải tác dụng 200g H2O C% dung dịch sau phản ứng
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+65b=12,1\) (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Bảo toàn electron: \(2n_{Fe}+2n_{Zn}=2n_{H_2}\) \(\Rightarrow2a+2b=0,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{12,1}\cdot100\%\approx46,28\%\\\%m_{Zn}=53,72\%\end{matrix}\right.\)
b)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=n_{Zn}=n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(p.ứ\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow\Sigma n_{H_2SO_4}=0,2\cdot110\%=0,22\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,22-0,2\right)\cdot98=1,96\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=211,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{211,7}\cdot100\%\approx7,18\%\\C\%_{ZnSO_4}=\dfrac{16,1}{211,7}\cdot100\%\approx7,61\%\\C\%_{H_2SO_4}=\dfrac{1,96}{22,4}\cdot100\%\approx0,93\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot9,8\%}{98}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\) \(\Rightarrow\) Fe2O3 còn dư, tính theo axit
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,1\cdot400=40\left(g\right)\\m_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\cdot160\approx5,3\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{40}{16+200-5,3}\cdot100\%\approx18,98\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(m_{ct}=\dfrac{19,6.200}{100}=39,2\left(g\right)\)
\(n_{H2SO4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
a) Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
0,2 0,4 0,1 0,3
b) Lap ti so so sanh : \(\dfrac{0,2}{2}< \dfrac{0,4}{3}\)
⇒ Al phan ung het , H2SO4 du
⇒ Tinh toan dua vao so mol cua Al
\(n_{Al2\left(SO4\right)3}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
\(n_{H2SO4\left(du\right)}=0,4-\left(\dfrac{0,2.3}{2}\right)=0,1\left(mol\right)\)
⇒ \(m_{H2SO4\left(du\right)}=0,1.98=9,8\left(g\right)\)
\(m_{ddspu}=5,4+200-\left(0,3.2\right)=204,8\left(g\right)\)
\(C_{Al2\left(So4\right)3}=\dfrac{34,2.100}{204,8}=16,7\)0/0
\(C_{H2SO4\left(du\right)}=\dfrac{9,8.100}{204,8}=4,78\)0/0
Chuc ban hoc tot
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
m_{H_2SO_4}=200.12,25\%=24,5g\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\
LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\)
\(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nMgO=0,15(mol); nH2SO4=0,4(mol)
PTHH: MgO + H2SO4 -> MgSO4 + H2O
0,15________0,15__________0,15(mol)
Ta có: 0,15/1 < 0,4/1
=> H2SO4 dư, MgO hết, tính theo nMgO
-> nH2SO4(dư)=0,4-0,15=0,25(mol) => mH2SO4(dư)=24,5(g)
nMgSO4=nMg=0,15(mol) => mMgSO4=120.0,15=18(g)
mddsau=6+200=206(g)
=>C%ddH2SO4(dư)=(24,5/206).100=11,893%
C%ddMgSO4=(18/206).100=8,738%
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot19,6\%}{98}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit dư
\(\Rightarrow\left\{{}\begin{matrix}n_{MgSO_4}=n_{H_2}=0,25\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgSO_4}=0,25\cdot120=30\left(g\right)\\m_{H_2}=0,25\cdot2=0,5\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,15\cdot98=14,7\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddH_2SO_4}-m_{H_2}=205,5\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{30}{205,5}\cdot100\%\approx14,6\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{14,7}{205,5}\cdot100\%\approx7,2\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề bài ta có : \(\left\{{}\begin{matrix}nNaOH=\dfrac{100.10}{100.40}=0,25\left(mol\right)\\nHNO3=\dfrac{200.6,3}{100.63}=0,2\left(mol\right)\end{matrix}\right.\)
Ta có PTHH :
\(NaOH+HNO3->NaNO3+H2O\)
0,2mol...........0,2mol.........0,2mol
a) Theo PTHH ta có : \(nNaOH=\dfrac{0,25}{1}mol>nHNO3=\dfrac{0,2}{1}mol\)
=> nNaOH dư
=> dd sau p/ư sẽ bao gồm NaNO3 và NaOH dư
Vậy khi cho quỳ tím vào dd sau p/ư thì quỳ tím sẽ hóa xanh .
b) Ta có : \(\left\{{}\begin{matrix}C\%NaOH\left(d\text{ư}\right)=\dfrac{\left(0,25-0,2\right).40}{\left(0,25-0,2\right).40+200}.100\%\approx0,99\%\\C\%NaNO3=\dfrac{0,2.85}{\left(0,25-0,2\right).40+200}.100\%\approx8,42\%\end{matrix}\right.\)
\(n_{Ba}=\dfrac{27.4}{137}=0.2\left(mol\right)\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
\(0.2........................0.2...........0.2\)
\(m_{Ba\left(OH\right)_2}=0.2\cdot171=34.2\left(g\right)\)
\(m_{dd_{Ba\left(OH\right)_2}}=27.4+200-0.2\cdot2=227\left(g\right)\)
\(C\%_{Ba\left(OH\right)_2}=\dfrac{34.2}{227}\cdot100\%=15.06\%\)
Số mol của bari
nBa = \(\dfrac{m_{Ba}}{M_{Ba}}=\dfrac{27,4}{137}=0,2\left(mol\right)\)
Pt : Ba + 2H2O \(\rightarrow\) Ba(OH)2 + H2 \(|\)
1 2 1 1
0,2
Số mol của bari hidroxit
nBa(OH)2 = \(\dfrac{0,2.1}{1}\) = 0,2 (mol)
Khối lượng của bari hidroxit
mBa(OH)2 = nBa(OH)2 . MBa(OH)2
= 0,2 . 171
= 34,2 (g)
Khối lượng của dung dịch
mdd = mct + mdm
= 34,2 + 200
= 234,2 (g)
Nồng độ phần trăm của dung dịch sau phản ứng
C0/0 Ba(OH) = \(\dfrac{m_{ct}.100}{m_{dd}}\) = \(\dfrac{34,2.100}{234,2}\)
= 14,6 0/0
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