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![](https://rs.olm.vn/images/avt/0.png?1311)
nAl= 0,5(mol)
a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
nHCl= 6/2 . 0,5= 1,5(mol)
=>mHCl= 1,5.36,5=54,75(mol)
=> mddHCl= (54,75.100)/18,25=300(g)
b) nH2= 3/2. 0,5=0,75(mol)
=>V(H2,đktc)=0,75.22,4=16,8(l)
c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)
mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)
=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$
b)
$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)
$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$
$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
0,2------------------>0,4---->0,2
mdd sau pư = 200 + 21,2 - 0,2.44 = 212,4(g)
=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)
$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$
$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$
Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$
$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$
$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow\)
Ta có: \(n_{SO_2}=\dfrac{2,479}{22,4}=\dfrac{2479}{22400}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=\dfrac{2479}{22400}\left(mol\right)\\n_{HCl}=\dfrac{2479}{11200}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_3}=\dfrac{2479}{22400}\cdot126\approx13,94\left(g\right)\\C_{M_{HCl}}=\dfrac{\dfrac{2479}{11200}}{0,25}\approx0,89\left(M\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)
\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)
\(n_{Fe}= \dfrac{5,6}{56}= 0,1 mol\)
Theo PTHH:
\(n_{H_2}= n_{Fe}= 0,1 mol\)
Vì ở điều kiện tiêu chuẩn(t độ=25 độ C, áp suất 1Bar), 1 mol khí chiếm thể tích 24,79 lít.
\(\Rightarrow V= 0,1 . 24,79= 2,479 l\)
b)
Theo PTHH:
\(n_{HCl}= 2n_{Fe}= 0,2 mol \Rightarrow m_{HCl}= 0,2 . 36,5=7,3 g\)
\(\Rightarrow m_{dd HCl}= \dfrac{ 7,3 . 100}{10}= 73 g\) (nếu bạn viết ở tử : 7,3 . 100% thì ở mẫu bạn phải viết là 10% nhé)
c) dd sau pư là FeCl2
Ta có: m dd sau pư= mFe + m dung dịch HCl - mH2
= 5,6 + 73 - 0,1 . 2=78,4 g
Theo PTHH:
\(n_{FeCl_2}= n_{Fe}= 0,1 mol \Rightarrow m_{FeCl_2}= 0,1 . 127= 12,7 g\)
C%FeCl2= \(\dfrac{12,7}{78,4}\).100% = 16,199%