Ta có: \(n_{H_2}=\dfrac{4,2}{22,4}=0,1875\left(mol\right)\)

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

__0,125__0,375___________0,1875 (mol)

\(\Rightarrow m_{Al}=0,125.27=3,375\left(g\right)\)

\(V_{HCl}=\dfrac{0,375}{3}=0,125\left(l\right)=125\left(ml\right)\)

Bạn tham khảo nhé!

n H2 = 4,2/22,4 = 3/16 mol

2Al + 6HCl $\to$ 2AlCl3 + 3H2

Theo PTHH :

n HCl = 2n H2 = 3/8 mol => V dd HCl = (3/8) / 3 = 0,125M

n Al = 2/3 n H2 = 0,125(mol) => m = 0,125.27  = 3,375(gam)

\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.n_{Al}=\dfrac{8,1}{27}=0,3mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

\(m_{HCl}=0,9.36,5=32,85g\\ c.V_{H_2}=0,45.24,79=11,1555l\)

a)2Al + 6HCl → 2AlCl3 + 3H2
     0,3    0,9                         0,45          
b)
nAl= \(\dfrac{8,1}{27}=0,3\)
=>mHCl= 0,9. 36,5 = 32,85
c)
=> VH2= 0,45.22,4= 10,08l

Có j k hiểu hỏi mình nhá

mai mình thi r giúp mình vs

\(a.2Al+6HCl->2AlCl_3+3H_2\)

        0,1.....................................0,1..........0,15

b. n Al = \(\frac{2,7}{27}=0,1mol\)

m AlCl3 = 0,1(27+35,5.3)=13,35 g

V H2 = 0,15.22,4=3,36 lít

cảm ơn bạn nha

fs

Đề cho tác dụng đủ với 250 ml chất gì bạn nhỉ?

\(\left\{{}\begin{matrix}Al\\Zn\end{matrix}\right.+HCl\rightarrow\left\{{}\begin{matrix}AlCl_3\\ZnCl_2\end{matrix}\right.+H_2\)

Bảo toàn nguyên tố H:

\(n_{HCl}=2n_{H_2}=2.\dfrac{6,72}{22,4}=0,6\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,6}{2}=0,3\left(l\right)\)

\(\left\{{}\begin{matrix}AlCl_3\\ZnCl_2\end{matrix}\right.+AgNO_3\rightarrow\left\{{}\begin{matrix}Al\left(NO_3\right)_3\\Zn\left(NO_3\right)_3\end{matrix}\right.+AgCl\downarrow\)

Bào toàn nguyên tố Cl:

\(n_{AgCl}=n_{HCl}=0,6\left(mol\right)\)

\(\Rightarrow m=m_{AgCl}=0,6.143,5=86,1\left(g\right)\)

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

X là khí Hidro

b) Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Al}=0,2mol\) 

\(\Rightarrow\%m_{Al}=\dfrac{0,2\cdot27}{8,64}\cdot100\%=62,5\%\) \(\Rightarrow\%m_{Cu}=37,5\%\)

c) Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\) 

\(\Rightarrow V_{HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\ b,V_{ddMgCl_2}=V_{ddHCl}=0,2\left(l\right)\\ n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ C_{MddMgCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)

a) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,15->0,3------------->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

b) \(C\%\left(dd.HCl\right)=\dfrac{0,3.36,5}{100}.100\%=10,95\%\)