Đề cho bao nhiêu gam axit clohidric bạn nhỉ?

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(n_{HCl}=\dfrac{3}{36,5}=\dfrac{6}{73}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{\dfrac{6}{73}}{2}\), ta được Zn dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{3}{73}\left(mol\right)\)

\(\Rightarrow V_{H_2}=\dfrac{3}{73}.22,4=\dfrac{336}{365}\left(l\right)\)

a )

n_Mg = \(\dfrac{7,2}{24}\) 0,3 ( mol )
Mg + H₂SO₄ -> MgSO₄ + H₂

Theo pt : m_mgSO4 = 0,3.120 = 36 ( g )

b ) 

Theo pt : n_H2 = n_Mg = 0,3 ( mol )

-> V_{H2( đktc )} = 0,3.22,4 = 6,72 ( l )

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a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,35}{1}\), ta được H₂ dư.

Theo PT: \(n_{H_2\left(pư\right)}=n_{CuO}=0,15\left(mol\right)\)

\(\Rightarrow n_{H_2\left(dư\right)}=0,35-0,15=0,2\left(mol\right)\)

\(\Rightarrow m_{H_2\left(dư\right)}=0,2.2=0,4\left(g\right)\)

\(n_{Fe}=\dfrac{28}{56}=0,5mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,5    1                           0,5

\(V_{H_2}=0,5\cdot22,4=11,2l\)

\(m_{HCl}=1\cdot36,5=36,5g\)

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

Làm gộp các phần còn lại

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

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

số mol kẽm tham gia phản ứng là:\(n_{Zn}=\frac{m}{M}=\frac{6,5}{65}=0,1\left(mol\right)\)

PTHH:

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1 0,2 0,1 (mol)

a, thể tích khí hiđro thu được là:\(V_{H_2}=n_{H_2}\times22,4=0,1\times22,4=2,24\left(l\right)\)

b,khối lượng HCl cần dùng là:\(m_{HCl}=n_{HCl}\times M=0,2\times65=13\left(g\right)\)

số mol kẽm tham gia phản ứng là:nZn=mM=6,565=0,1(mol)nZn=mM=6,565=0,1(mol)

PTHH:

0,1 0,2 0,1 (mol)

a):VH2=nH2×22,4=0,1×22,4=2,24(l)VH2=nH2×22,4=0,1×22,4=2,24(l)

b):mHCl=nHCl×M=0,2×65=13(g)

a, Ta có pt : \(Zn+2HCL->ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo pt (ở câu a ) , có :  \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(V_{H_2\left(DKTC\right)}=0,1\times22,4=2,24\left(l\right)\)

n_Zn= \(\dfrac{6,5}{65}\) = 0,1 (mol)
a. PTHH: Zn + 2HCl -> ZnCl₂ + H₂

               0,1              -> 0,1 -> 0,1  (mol) 
b. n_H2 = 0,1 x 22,4 = 2,24 (l)
 

a. PTHH: Fe + 2HCl ---> FeCl₂ + H₂ (1)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo pthh (1): \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

\(\rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, PTHH: 2H₂ + O₂ --t^o--> 2H₂O (2)

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

\(\rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)