\(300(ml)=0,3(l)\\ n_{HCl}=1.0,3=0,3(mol);n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ \text{LTL: }\dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,3-0,1.2=0,1(mol)\\ \Rightarrow m_{HCl(dư)}=0,1.36,5=3,65(g)\\ b,n_{FeCl_2}=n_{Fe}=0,1(mol)\\ \Rightarrow \begin{cases} C_{M_{FeCl_2}}=\dfrac{0,1}{0,3}=0,33M\\ C_{M_{HCl(dư)}}=\dfrac{0,1}{0,3}=0,33M \end{cases}\)

 a) nFe= 16/56 =~ 0,3 mol 

mH2S04 =( C% .mdd ) /100%= ( 20.100) /100 = 20g 

nH2SO4 = 20/98 =~ 0,2mol  

lập pthh của pu 

Fe + H2SO4 ----------> FeSO4 + H2

1mol   1mol                 1mol        1mol 

0,3mol  0,2mol 

xét tỉ lệ  nFe dư sau pư vậy tính theo mol H2SO4 

nFe (pư) = (0,2 .1 )/1 =0,2mol 

nFe (dư) = 0,3 -0,2 =0,1mol 

mFe dư = 0,1 . 56 = 5,6 g

mFeSO4 = 0,2 .152 = 30,4 g 

b) mdd sau pư = mFe + m dung môi  = 16 +100=116 g

c% Fe = (5,6 / 116) .100%=~ 4,83%

c% FeSO4 =(30,4/116).100%=~ 26,21%

a)  đối 200ml =0,2 lít 

 CMFe =n/v = 0,1 / 0,2 =0,5 mol/lít 

    CMFeSO4 =n/v = 0,2/0,2=1 mol /lít  

a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:        0,2         0,1            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H₂SO₄ dư

b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)

c/ V_{dd sau pứ} = 0,2 + 0,3 = 0,5 (l)

d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)

 \(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)

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

\(0.2......0.4..........0.2........0.2\)

\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

\(n_{FeSO_4}=0,5.0,3=0,15\left(mol\right)\\ n_{Mn}=\dfrac{5,5}{55}=0,1\left(mol\right)\\ Mn+FeSO_4\rightarrow MnSO_4+Fe\\ Vì:\dfrac{0,15}{1}< \dfrac{0,1}{1}\Rightarrow FeSO_4dư\\ n_{MnSO_4}=n_{Mn}=0,1\left(mol\right)\\ n_{FeSO_4\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\\ V_{ddsau}=V_{ddFeSO_4}=0,3\left(l\right)\\ C_{MddFeSO_4\left(dư\right)}=\dfrac{0,05}{0,3}=\dfrac{1}{6}\left(M\right)\\ C_{MddMnSO_4}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

C% thì em bổ sung KLR của dd nhé!

n_HCl = 0.4 (mol) 

n_Ba(OH)2 = 0.15 (mol) 

Ba(OH)₂ + 2HCl => BaCl₂ + H₂O

0.15_______0.3_____0.15 

C_{M HCl(dư)} = 0.3/0.4 = 0.75M

C_{M BaCl2} = 0.15/(0.4 + 0.3) = 3/14 (M) 

n_Fe = 0,1 mol

n_HCl = 0,3 mol

Fe + 2HCl ---> FeCl₂ + H₂

0,1 < 0,3/2 .....=> HCl dư sau phản ứng

n_FeCl2 = 0,1 mol => C_M = 0,1/0,2 = 0,5M

n_HCl(dư) = 0,1 mol => C_M = 0,1/0,2 = 0,5M

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=0,2\cdot1,5=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) Fe p/ứ hết, HCl còn dư

\(\Rightarrow n_{HCl\left(dư\right)}=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)

c) Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)

\(\Rightarrow C_{M_{FeCl_2}}=C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)