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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,4 1,2 0,4 0,6
\(a,V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)
\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)
\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)
\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)\)
\(n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
trc p/u : 0,3 0,75
p/u: 0,3 0,6 0,3 0,3
sau p/u : 0 0,15 0,3 0,3
---> Sau p/ư HCl dư
\(a,V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(b,m_{ddHCl}=0,6.36,5=21,9\left(g\right)\)
\(c,m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
\(m_{ddZnCl_2}=19,5+150-\left(0,3.2\right)=168,9\left(g\right)\)
\(C\%=\dfrac{40,8}{168,9}.100\%\approx24,16\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a) PTHH: Fe + 2HCl --> FeCl2 + H2
_______0,2---->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48(l)
b) mHCl = 0,4.36,5 = 14,6(g)
c) mFeCl2 = 0,2.127 = 25,4 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl = 5.4 / 27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2......0.6............0.2.......0.3
a) VH2 = 0.3 * 22.4 = 6.72 (l)
b) mAlCl3 = 0.2 * 133.5 = 26.7 (g)
c) VddHCl = 0.6 / 1.5 = 0.4 (l)
d) CMAlCl3 = 0.2 / 0.4 = 0.5 (M)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(pthh:Mg+2HCl--->MgCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\n_{HCl}=\dfrac{\dfrac{109,5.10\%}{100\%}}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)
a. Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\)
Vậy Mg dư
Theo pt: \(n_{H_2}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,15.2=0,3\left(g\right)\)
b. \(m_{dd_{sau.PỨ}}=4,8+109,5-0,3=114\left(g\right)\)
c. Theo pt: \(n_{MgCl_2}=0,15\left(mol\right)\)
\(\Rightarrow C_{\%_{MgCl_2}}=\dfrac{0,15.95}{114}.100\%=12,5\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
n Mg=\(\dfrac{4,8}{24}\)=0,2 mol
m HCl=\(\dfrac{109,5.10}{100}\)=10,95g
=>n HCl=\(\dfrac{10,95}{36,5}\)=0,3 mol
Mg + 2HCl -> MgCl2 + H2
0,15---0,3---------0,15----0,15 mol
=>HCl phản ứng hết, Mg dư
=> mMg = 0,15.24 = 3,6 (g)
=> mH2 = 0,15.2 = 0,3 (g)
mddspu= mMg + mddHCl- mH2 = 3,6 + 109,5 - 0,3 = 112,8 (g)
c)
mMgCl2= 0,15.95 = 14,25 (g)
->C%ddMgCl2=\(\dfrac{14,25}{112,8}\).100=12,63%
cho e hỏi là sao ở tên tính nMg ra 0,2 sao ở dưới lại viết là 0,15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(m_{HCl}=\dfrac{109,5\cdot10\%}{100\%}=10,95g\Rightarrow n_{HCl}=0,3mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,1 0,3 0 0
0,1 0,2 0,1 0,1
0 0,1 0,1 0,1
a)\(V_{H_2}=0,1\cdot22,4=2,24l\)
b)\(m_{MgCl_2}=0,1\cdot95=9,5g\)
c)\(m_{H_2}=0,1\cdot2=0,2g\)
\(m_{ddMgCl_2}=4,8+109,5-0,2=114,1g\)
\(C\%=\dfrac{9,5}{114,1}\cdot1005=8,33\%\)
\(a.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.10\%}{36,5}=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ LTL:\dfrac{0,2}{1}>\dfrac{0,3}{2}\\ \Rightarrow Mgdư\\ n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{H_2}=0,15.2=0,3\left(g\right)\\ b.m_{ddsaupu}=m_{Mg\left(pứ\right)}+m_{ddHCl}-m_{H_2}=0,15.24+109,5-0,3=112,8\left(g\right)\\ c.n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,15.95}{112,8}.100=12,63\%\)
a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
n Al = 5,4/27 = 0,2(mol)
Theo PTHH : n H2 = 3/2 n Al = 0,3(mol)
=> V H2 = 0,3.22,4 = 6,72(lít)
b) n HCl = 3n Al = 0,6(mol)
=> mdd HCl = 0,6.36,5/20% = 109,5 gam
c)Sau phản ứng,
mdd = m Al + mdd HCl - m H2 = 5,4 + 109,5 - 0,3.2 = 114,3(gam)
=> C% AlCl3 = 0,2.133,5/114,3 .100% = 23,36%
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