\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)

\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ a,PTHH:CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ b,n_{CO_2}=n_{\left(CH_3COO\right)_2Ca}=n_{CaCO_3}=0,2\left(mol\right)\\ b,V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Ca}=0,2.158=31,6\left(g\right)\\ d,C_4H_{10}+\dfrac{5}{2}O_2\rightarrow2CH_3COOH+H_2O\\ n_{C_4H_{10}\left(LT\right)}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ n_{C_4H_{10}\left(TT\right)}=0,2:50\%=0,4\left(mol\right)\\ m_{C_4H_{10}\left(tt\right)}=58.0,4=23,2\left(g\right)\)

\(n_{BaCO_3}=\dfrac{19,7}{197}=0,1\left(mol\right)\\ a,PTHH:BaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ba+CO_2+H_2O\\ n_{CO_2}=n_{\left(CH_3COO\right)_2Ba}=n_{BaCO_3}=0,1\left(mol\right)\\ b,V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Ba}=255.0,1=25,5\left(g\right)\\ d,C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ n_{CH_3COOH}=0,1.2=0,2\left(mol\right);n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ n_{C_2H_5OH\left(TT\right)}=0,2.75\%=0,15\left(mol\right)\\ m_{C_2H_5OH\left(TT\right)}=0,15.46=6,9\left(g\right)\)

nMg = 3,6 : 24 = 0,15 (mol) 
pthh : Mg + 2HCl --> MgCl2 + H2 
           0,15-----------> 0,15 --->0,15 (mol) 
mMgCl2 = 0,15 . 95 = 14,25 (mol) 
VH2 (đkc)= 0,15. 24,79 = 3,718(l) 
 

\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,15      0,3         0,15       0,15

\(m_{MgCl_2}=0,15\cdot95=14,25g\)

\(V_{H_2}=0,15\cdot22,4=3,36l\)

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

n H₂ = \(\dfrac{6,72}{22,4}=0,3\) (mol)

=> n Zn = n H₂ = 0,3 mol

=> m Zn = 0,3.65= 19,5 (g)

a) 

Zn  +  2HCl  →  ZnCl₂   +  H₂ 

b) nZn = \(\dfrac{3,5}{65}\)=\(\dfrac{7}{130}\) mol 

Theo tỉ lệ phản ứng => nH₂ = nZn= \(\dfrac{7}{130}\)mol

<=> V H₂ = \(\dfrac{7}{130}\).22,4 = 1,206 lít

c) nZnCl₂ = nZn => mZnCl₂ = \(\dfrac{7}{130}\).136= 7,32 gam

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

b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)  

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,2`   `0,4`                `0,2`      `0,2`       `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`b)V_[H_2]=0,2.22,4=4,48(l)`

`c)m_[dd HCl]=[0,4.36,5]/5 . 100=292(g)`

`=>C%_[ZnCl_2]=[0,2.136]/[13+292-0,2.2].100~~8,93%`

Zn + 2HCl → ZnCl₂ + H₂

0,2-----0,4------0,2------0,2

n Zn=0,2 mol

=>VH2=0,2.22,4=4,48l

=>mddHcl=292g

C%=\(\dfrac{0,2.136}{13+292-0,2.2}100=8,929\%\)

a. Zn (0,2 mol) + 2HCl (0,4 mol) \(\rightarrow\) ZnCl₂ (0,2 mol) + H₂ (0,2 mol).

b. Số mol Zn là 13:65=0,2 (mol).

Thể tích khí H₂ thu được ở đktc là 0,2.22,4=4,48 (l).

c. Khối lượng dung dịch là 0,4.36,5.100:5+13-0,2.2=304,6 (g).

Nồng độ phần trăm cần tìm là 0,2.136/304,6.100%\(\approx\)8,93%.

Sửa đề : 11.2 g sắt

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

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

\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)

\(m_{Fe}=\dfrac{11}{56}=0.19\left(mol\right)\)

\(Fe+H_2SO_4->FeSO_4+H_2\)

0.19   0.19            0.19              (mol)

\(m_{FeSO_4}=0.19\cdot152=28.28\left(g\right)\)

\(Cm=\dfrac{0.19}{0.05}=3.8M\)

nFe=11.256=0.2(mol)

mFeSO4=0,2⋅152=30,4(g)