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)

a)

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

                 1,3<---4<-------1,3<---------2                      

b)

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

\(m_{AlCl_3}=n\cdot M=1,3\cdot\left(27+35,5\cdot3\right)=173,55\left(g\right)\)

\(m_{Al}=n\cdot M=1,3\cdot27=35,1\left(g\right)\)

a) 2Al + 6HCl → 2AlCl3 + 3H2

b) n Al= 8,1/27 =0,3(mol)

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4= 10,08 lít

c) n AlCl3 = n Al = 0,3 mol

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 2n H2 = 0,9(mol)

=> V dd HCl = 0,9/0,5 = 1,8(lít)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,1(mol);n_{HCl}=0,3(mol)\\ b,m_{Al}=0,1.27=2,7(g);m_{HCl}=0,3.36,5=10,95(g)\\ m_{AlCl_3}=0,1.133,5=13,35(g)\\ c,n_{Al}=\dfrac{16,2}{27}=0,6(mol)\\ \Rightarrow n_{H_2}=1,5n_{Al}=0,9(mol)\\ \Rightarrow V_{H_2}=0,9.22,4=20,16(l)\)

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

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)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2----------->0,2----->0,3

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

                      0,3<----------------0,3

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH:

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

0,2--------------->0,2------->0,3

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c, PTHH:

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

           0,2<------------------0,2

\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

nAl = 8,1 /27 = 0,3mol

2Al + 6HCl => 2AlCl3 + 3H2

0,3--------------->0,3------> 0,45

=> VH2 = 0,45.22,4 = 10,08 (l)

mAlCl3 = 0,3. 133,5 = 40,05 (g)