2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g

B nha

a. 2Al + 6HCl -> 2AlCl₃ + 3H₂

b. n_Al = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)

\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)

c. \(n_{HCl}=3n_{Al}=3.0,3=0,9\left(mol\right)=>m_{HCl}=0,9.36,5=32,85\left(g\right)\)

Vậy m = 32,85

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.2.........................0.2.......0.3\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

\(n_{CuO}=\dfrac{32}{160}=0.2\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1..........1\)

\(0.2........0.3\)

\(LTL:\dfrac{0.2}{1}< \dfrac{0.3}{1}\Rightarrow H_2dư\)

\(n_{Cu}=0.2\left(mol\right)\)

\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)

Em xem lại đề vì chất rắn chỉ có Cu không có CuO nhé !

a) Nhôm + Axit clohidric --> Nhôm clorua + Khí hidro

b) 2Al + 6HCl --> 2AlCl₃ + 3H₂

c) Theo ĐLBTKL: m_Al + m_HCl = m_AlCl3 + m_H2

=> m_HCl = 26,7 + 0,6 - 5,4 = 21,9(g)

 PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\) 

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b)

\(CM=\dfrac{n}{v}\Leftrightarrow n=v.CM=1,5.0,6=0,9molHCl\)

\(0,9molHCl\rightarrow\dfrac{0,9.2}{6}=0,3molAl\)

m=n.M=0,3.27=8,1g

Trả lời giúp em với plsss🥺

a) PTHH : 2Al + 6HCl -------------->  2AlCl3 + 3H2

b ) b. Tính khối lượng kim loại Al cần dùng..

\(n_{HCl}=1,5.0,6=0,9\left(mol\right)\)

Theo PT : n_Al = \(\dfrac{1}{3}n_{HCl}\)=0,3(mol)

=> m _Al= 0,3.27 = 8,1(g)

Chúc em học tốt !!!

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

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

nH₂=6.72/22,4=0,3(mol)
PTHH: 2Al +   6HCl --> 2AlCl₃ + 3H₂
bài ra:  0,2 <-- 0,6  <--   0,2   <-- 0,3  /mol
a) mHCl = 0,6.36,5=21,9(g)
b) mAl = 0.2.24 = 4,8(g)
 

\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)

              0,2<--0,6<----------0,2<------0,3    (mol)

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)

\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)