a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, Phần này có lẽ đề cho Sắt (III) oxit bạn nhỉ?

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\), ta được Fe₂O₃ dư.

Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(g\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

a+b+c) Ta có: \(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,25\left(mol\right)=n_{MgCl_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,25\cdot95=23,75\left(g\right)\\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.\)

d) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

Theo PTHH: \(n_{Cu}=n_{H_2}=0,25\left(mol\right)\) \(\Rightarrow m_{Cu}=0,25\cdot64=16\left(g\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{ZnCl_2}=\dfrac{13,6}{136}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,1<-----------0,1----->0,1

=> m_Zn = 0,1.65 = 6,5 (g)

b) V_H2 = 0,1.22,4 = 2,24 (l)

a:

nFe=11,2/56=0,2(mol)

Fe+2HCl->FeCl2+H2

0,2                           0,2

b: V=0,2*22,4=4,48(lít)

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow H_2hết,CuOdư\\ n_{Cu}=n_{CuO\left(P.Ứ\right)}=n_{H_2}=0,2\left(mol\right)\\ n_{CuO\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ m_{rắn}=m_{Cu}+m_{CuO\left(DƯ\right)}=0,2.64+0,1.80=20,8\left(g\right)\)

- PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

- Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

- PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,075\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,075.22,4=1,68\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=8,4\left(l\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ TheoPT:n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\ b.2H_2+O_2-^{t^o}\rightarrow2H_2O\\ n_{O_2}=\dfrac{1}{2}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)

Ý a , b , riêng nhé 

Giúp mình với ạ mình cần gấp

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

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

Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

Khối lượng muối thu được:

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

b. Theo PTHH: \(n_{H_2}=\dfrac{n_{Al}.3}{2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

Thể tích khí hiđro thu được:

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72\left(l\right)\)

a) \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\) 

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\) 

\(n_{HCl}=2.n_{Mg}=0,2.2=0,4mol\)

\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6g\)

c) \(n_{H_2}=n_{Mg}=0,2mol\) 

Thể tích khí hidro sinh ra (ở đktc): 

\(V_{H_2}=0,2.24,79=4,958l.\)

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

0,4--------------------------0,6

n Al=0,4 mol

=>VH2=0,6.22,4=13,44l

H2+HgO-tO>Hg+H2O

0,6--------------0,6

=>m Hg=0,6.201=120,6g

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)

                0,4                                    0,6

\(\rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\\ PTHH:HgO+H_2\underrightarrow{t^o}Hg+H_2P\)

                           0,6    0,6

\(\rightarrow m_{Hg}=0,6.201=120,6\left(g\right)\)

nAl = 10,8 : 27 = 0,4  (mol) 
pthh : 2Al + 6HCl ---> 2AlCl3 + 3H2
          0,4---------------------------->0,6 (mol) 
=> VH2 = 0,6 . 22,4 = 13,44 (L)
pthh : HgO + H2 -t--> Hg + H2O 
          0,6<---0,6------>0,6-->0,6 (MOL) 
=> mHg = 0,6 . 201 =120,6 (G)