al td voi cl tao khi tinh mol khi suy he pt tinh khoi luong al lay khoi luong hh - khoi luong al tinh duoc khoi luong nhom oxit sau do tinh thanh phan phan tram.b)tu im

2 so mol cua 2 hon hop suy he pt tinh mol cua hcl tinh m hcl sau do tinh mddhcl dung cong thuc v=m/d tinh duoc vdd can t

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có :

\(n_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)

\(\rightarrow n_{AL}=\frac{2}{3}n_{H2}=0,4\left(mol\right)\)

\(\rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)

\(\rightarrow\%m_{Al}=\frac{10,8}{21}.100\%=51,43\%\)

\(\%m_{Al2O3}=100\%-51,43\%=48,57\%\)

b, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(n_{HCl}=3n_{Al}+6n_{Al2O3}=1,8\left(mol\right)\)

\(\rightarrow m_{HCl}=18.36,5=65,7\left(g\right)\)

\(\rightarrow m_{dd_{HCl}}=\frac{65,7}{36}.100\%=182,5\left(g\right)\)

\(\rightarrow V_{HCl}=\frac{182,5}{1,18}=154,66\left(l\right)\)

PTHH: 2Al + 6HCl ===> 2AlCl₃ + 3H₂ (1)

0,4 1,2 0,6 (mol)

Al₂O₃ + 6HCl ===> 2AlCl₃ + 3H₂O (2)

0,1 0,6 (mol)

a/ 13,44 lít khí chính là sản phẩm của Al tác dụng với HCl

n_H2 = 13,44 / 22,4 = 0,6 mol

Lập các sô mol trên phương trình, ta có

m_Al = 0,4 x 27 = 10,8 gam

m_Al2O3 = 21 - 10,8 = 10,2

=> Tính %m ( bạn tự tính nha)

b/ n_Al2O3 = 10,2 / 102 = 0,1 mol

=> n_HCl (2) = 0,1 x 6 = 0,6 mol

Lại có : n_HCl (1) = 1,2 mol

=> Tổng số mol HCl = 0,6 + 1,2 = 1,8 mol

=> m_HCl = 1,8 x 36,5 = 65,7 gam

=> m_{dung dịch HCl 36%} = \(\frac{65,7.100}{36}\) = 182,5 gam

=> V_HCl = 182,5 / 1,18 = 154,66 ml

2Al + 6HCl → 2AlCl₃ + 3H₂↑ (1)

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (2)

\(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)

a) Theo pt1: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,6=0,4\left(mol\right)\)

\(\Rightarrow m_{Al}=0,4\times27=10,8\left(g\right)\)

\(\Rightarrow m_{Al_2O_3}=21-10,8=10,2\left(g\right)\)

b) Theo PT1: \(n_{HCl}=3n_{Al}=3\times0,2=0,6\left(mol\right)\)

\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)

Theo Pt2: \(n_{HCl}=6n_{Al_2O_3}=6\times0,1=0,6\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,6+0,6=1,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,2\times36,5=43,8\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{43,8}{36\%}=121,67\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\frac{121,67}{1,18}=103,11\left(ml\right)\)

ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g 

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

Gọi số mol của Na₂CO₃ là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)

Gọi số mol của CaCO₃ là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)

n H2 = 13,44/22,4 = 0,6(mol)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$

n Al = 2/3 n H2 = 0,4(mol)

%m Al = 0,4.27/21  .100% = 51,43%
%m Al2O3 = 100% -51,43% = 48,57%

b)

=> n Al2O3 = (21 - 0,4.27)/102 = 0,1(mol)

$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$

n HCl = 3n Al + 6n HCl = 0,4.3 + 0,1.6 = 1,8(mol)

=> m dd HCl = 1,8.36,5/20% = 328,5(gam)

=> V dd HCl = m/D = 328,5/1,18 = 278,39(ml)

\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)

\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)

\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

Fe + 2HCl --> FeCl₂ + H₂

b) Gọi số mol Al, Fe lần lượt là a,b 

=> 27a + 56b = 13,9

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

a----->3a--------->a------->1,5a______(mol)

Fe + 2HCl --> FeCl₂ + H₂

b------>2b-------->b----->b__________(mol)

=> 1,5a + b = 0,35

=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c) n_HCl = 3a + 2b = 0,7 (mol)

=> m_HCl = 0,7.36,5 = 25,55(g)

=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)

\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)