Bài 3. Tính chất hóa học của axit

Gọi x và y lần lượt là số mol của Fe và Ca

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

x            x             x             x  (mol)

\(Ca+H_2SO_4\rightarrow CaSO_4+H_2\)

y          y                 y          y   (mol)

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Ta có hệ pt:

\(\left\{{}\begin{matrix}56x+40y=1,84\\x+y=0,04\end{matrix}\right.\)

=> x = 0,015

y = 0,025

=> mFe = 56x = 56 . 0,015 = 0,84(g)

mCa = 40y = 40 . 0,025 = 1 (g)

Zn+H2SO4->ZnSO4+H2

0,1---0,1-----------------0,1

n H2=0,1 mol

=>m Cu=10,5-0,1.65=4g

bài 4

CO2+Ba(OH)2->BaCO3+H2O

0,1------0,1-----------0,1

n CO2=0,1 mol

=>CM Ba(OH)2=\(\dfrac{0,1}{0,2}=0,5M\)

=>m BaCO3=0,1.197=19,7g

\(n_{K_2O}=\dfrac{28,2}{94}=0,3\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{230.9,8\%}{98}=0,23\left(mol\right)\)

PTHH: K₂O + H₂SO₄ --> K₂SO₄ + H₂O

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,23}{1}\) => H₂SO₄ hết, K₂O dư và pư với nước

PTHH: K₂O + H₂SO₄ --> K₂SO₄ + H₂O

           0,23<----0,23------>0,23

            K₂O + H₂O --> 2KOH

            0,07----------->0,14

m_{dd sau pư} = 28,2 + 230 = 258,2 (g) 

\(\left\{{}\begin{matrix}m_{K_2SO_4}=0,23.174=40,02\left(g\right)\\m_{KOH}=0,14.56=7,84\left(g\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{40,02}{258,2}.100\%=15,5\%\\C\%_{KOH}=\dfrac{7,84}{258,2}.100\%=3,04\%\end{matrix}\right.\)

\(n_{HCl}=0,1.0,5=0,05\left(mol\right)\)

PTHH: Ba(OH)₂ + 2HCl --> BaCl₂ + 2H₂O

             0,025<----0,05---->0,025

=> \(V=\dfrac{0,025}{2}=0,0125\left(l\right)\)

\(C_{M\left(A\right)}=\dfrac{0,025}{0,0125+0,1}=\dfrac{2}{9}M\)

100ml = 0,1 (l) 

VHCl = 0,1 (l)

=> nHCl = CM.V = 0,1.0,5 = 0,05 (mol)

Ba(OH)₂ + 2HCl -- > BaCl₂ + 2H₂O

0,025         0,05          0,025     0,05 

VBa(OH)2 = n/CM = 0,025 / 2 = 0,0125 (l)

CMBaCl₂ = n/V = 0,025 / 0,0125 = 2M

Gọi số mol Al, Cu là a, b (mol)

Có: \(\left\{{}\begin{matrix}27a+64b=15,5\\a:b=1:2\end{matrix}\right.\)

=> a = 0,1 (mol); b = 0,2 (mol)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,1-->0,3------------>0,15

=> \(V=\dfrac{0,3}{1}=0,3\left(l\right)\)

a = 0,15.2 = 0,3 (g)

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

              0,1----->0,3------------>0,1

=> m = 0,1.342 = 34,2 (g)

\(V=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

\(n_{HCl}=10.0,05=0,5\left(mol\right)\)

PTHH: Fe₃O₄ + 8HCl --> FeCl₂ + 2FeCl₃ + 4H₂O

           0,0625<--0,5---->0,0625-->0,125

=> \(m_{Fe_3O_4}=0,0625.232=14,5\left(g\right)\)

\(\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,0625}{10}=0,00625M\\C_{M\left(FeCl_3\right)}=\dfrac{0,125}{10}=0,0125M\end{matrix}\right.\)

\(a,Zn+H_2SO_4\to ZnSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow \begin{cases} 65n_{Zn}+27n_{Al}=3,68\\ n_{Zn}+1,5n_{Al}=0,1 \end{cases}\\\Rightarrow n_{Al}=n_{Zn}=0,04(mol)\\ \Rightarrow \%m_{Al}=\dfrac{0,04.27}{3,68}.100\%=29,345\&\\ \Rightarrow \%m_{Zn}=100-29,34=70,66\%\\ b,n_{H_2SO_4}=n_{H_2}=0,1\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,1.98}{10\%}=98(g)\)

\(c,n_{ZnSO_4}=0,04(mol);n_{Al_2(SO_4)_3}=0,02(mol)\\ \Rightarrow m_{ZnSO_4}=161.0,04=6,44(g)\\ m_{Al_2(SO_4)_3}=0,02.342=6,84(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{6,84}{3,68+98-0,1.2}.100\%=6,74\%\\ \Rightarrow C\%_{ZnSO_4}=\dfrac{6,44}{3,68+98-0,1.2}.100\%=6,35\%\)

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

2a        3a                  a                3a

Zn + H2SO4 -> ZnSO4 + H2

b           b                b           b

nH2 = 0.1 mol

Ta có: \(\left\{{}\begin{matrix}54a+65b=3.68\\3a+b=0.1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.02mol\\b=0.04mol\end{matrix}\right.\)

a.\(\%mAl=\dfrac{0.02\times54\times100}{3.68}=29.35\%\)

%mZn = 100 - 29.35 = 70.65%

b.mddH2SO4\(=\dfrac{0.1\times98\times100}{10}=98g\)

Định luật bảo toàn khối lượng:

mKL + mdd ban đầu = mdd sau + mH2

<=> mdd sau = 3.68 + 98 - \(0.1\times2=101.48g\)

c.C%Al2(SO4)3 =\(\dfrac{\left(0.02\times342\right)\times100}{101.48}=6.74\%\) 

C%ZnSO4 = \(\dfrac{0.04\times161\times100}{101.48}=6.35\%\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(Fe_2\left(SO_4\right)_3+3BaCl_2\rightarrow2FeCl_3+3BaSO_4\downarrow\)

\(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Zn}=0,15(mol)\Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{25,95}.100\%=37,57\%\\ \Rightarrow \%_{ZnO}=(100-37,57)\%=62,43\%\\ c,n_{ZnO}=\dfrac{25,95-9,75}{81}=0,2(mol)\\ \Rightarrow n_{HCl}=2.0,15+2.0,2=0,7(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,7.36,5}{12\%}=212,92(g)\)