cho 8,4g bột sắt tác dụng với 200ml dd HCl 2M ( D= 1,25g/ml), thu được 200ml dung dịch A và V ml khí hidro (dktc)
a) tính V (ml)
b) tính nồng độ mol các chất tan trong dung dịch A
c) tính nồng độ % của chất tan trong dd A
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1---->0,1----------------->0,1
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\end{matrix}\right.\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 0,25 > 0,1 => CuO dư
Theo pthh: nCu = nH2 = 0,1 (mol)
=> mCu = 0,1.64 = 6,4 (g)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48l\\
C_M=\dfrac{0,2}{0,2}=1M\\
n_{CuO}=\dfrac{20}{80}=0,25\left(G\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:0,25>0,1\)
=>CuO dư
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\\
m_{Cu}=0,1.64=6,4g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,2(mol)\\ a,m_{BaSO_4}=0,2.233=46,6(g)\\ b,V_{dd_{BaCl_2}}=\dfrac{0,2}{1,5}\approx 0,13(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2+0,13}\approx 1,21M\)
\(d,\) Dd sau p/ứ là HCl nên làm quỳ tím hóa đỏ
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ n_{BaCl_2}=n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\\ b,V_{\text{dd}BaCl_2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)\\ c,C_{M\text{dd}HCl}=\dfrac{0,4}{\dfrac{2}{15}+0,2}=1,2\left(M\right)\\ d,V\text{ì}.c\text{ó}.\text{dd}.HCl\Rightarrow Qu\text{ỳ}.ho\text{á}.\text{đ}\text{ỏ}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Coi V dd HCl = 100(ml)
m dd HCl = 1,25.100 = 125(gam)
n HCl = 125.7,3%/36,5 = 0,25(mol)
[H+ ] = [Cl- ] = CM HCl = 0,25/0,1 = 2,5M
b)
n Al = 0,235(mol)
2Al + 6HCl $\to$ 2AlCl3 + 3H2
n HCl pư = 3n Al = 0,705(mol)
n HCl dư = 0,4.2 - 0,705 = 0,095(mol)
[H+ ] = CM HCl dư = 0,095/0,4 = 0,2375M
pH = -log([H+ ]) = 0,624
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: R + 2HCl --> RCl2 + H2
____0,15<-----------------0,15
=> \(M_R=\dfrac{3,6}{0,15}=24\left(Mg\right)\)
b)
PTHH: Mg + 2HCl --> MgCl2 + H2
__________0,3<-----0,15<---0,15
=> \(V=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)
\(C_{M\left(MgCl_2\right)}=\dfrac{0,15}{0,15}=1M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:R+2HCl\rightarrow RCl_2+H_2\\ \Rightarrow n_R=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow M_R=\dfrac{m_R}{n_R}=\dfrac{3,6}{0,15}=24\\ \)
\(\Rightarrow R\) là \(Magie\left(Mg\right)\)
\(b.n_{HCl}=2.n_{H_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{n_{HCl}}{C_M}=\dfrac{0,3}{2}=0,15\left(l\right)=150ml\)
\(n_{MgCl_2}=n_{Mg}=0,15\left(mol\right)\\ \Rightarrow C_{M_{ddMgCl_2}}=\dfrac{n_{MgCl_2}}{V}=\dfrac{0,15}{0,15}=1\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a.
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(V_{H_2}=24,79.0,2=4,958\left(l\right)\)
b.
\(n_{HCl}=2.n_{Fe}=0,4\left(mol\right)\\ CM_{HCl}=\dfrac{0,4}{0,2}=2M\)
a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)
c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)
⇒ m dd sau pư = 8,4 + 250 - 0,15.2 = 258,1 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)