Bài 3:

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right);n_{H_2SO_4}=\dfrac{300.9,8\%}{98}=0,3\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\Rightarrow H_2SO_4dư\\ ddA:H_2SO_{4\left(dư\right)},CuSO_4\\ n_{CuSO_4}=n_{CuO}=0,2\left(mol\right);n_{H_2SO_4\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ m_{ddsau}=16+300=316\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{0,2.160}{316}.100\approx10,127\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,1.98}{316}.100\approx3,101\%\)

Bài 2:

\(Đặt:n_{CuO}=a\left(mol\right);n_{Fe_2O_3}=b\left(mol\right)\left(a,b>0\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ \Rightarrow\left\{{}\begin{matrix}80a+160b=32\\2a+6b=0,5.2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\a,\%m_{CuO}=\dfrac{0,2.80}{32}.100\%=50\%\Rightarrow \%m_{Fe_2O_3}=100\%-50\%=50\%\\ b,CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{H_2SO_4}=a+3b=0,5\left(mol\right)\\ \Rightarrow m_{ddH_2SO_4}=\dfrac{0,5.98.100}{10}=490\left(g\right)\)

a)

M: Fe

B₁: Fe₂(SO₄)₃ 

B₂: SO₂

A₁: FeCl₂

A₂: FeCl₃

C₁: Cu(NO₃)₂

C₂: AgCl

D₁: Cu(OH)₂

D₂: NaNO₃

E₂: C₆H₁₂O₆

E₃: C₆H₁₂O₇

E₄: Ag

M₁: Fe₃O₄

M₂: FeO
M₃: CO₂

E₁: (C₆H₁₀O₅)_n 

a) \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

b) \(Fe+2HCl\rightarrow FeCl_2+H_2\)

c) \(2FeCl_2+Cl_2\xrightarrow[]{t^o}2FeCl_3\)

d) \(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\)

e) \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaNO_3\)

k) \(C_6H_{12}O_6+Ag_2O\xrightarrow[]{NH_3}C_6H_{12}O_7+2Ag\downarrow\)

f) \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

g) \(CO+Fe_3O_4\xrightarrow[]{t^o}3FeO+CO_2\)

h) \(6nCO_2+5nH_2O\xrightarrow[\text{chất diệp lục}]{\text{ánh sáng}}\left(C_6H_{10}O_5\right)_n\)

i) \(\left(C_6H_{10}O_5\right)_n+nH_2O\xrightarrow[]{axit}nC_6H_{12}O_6\)

\(X:Cu\left(NO_3\right)_2\\ Y:CuSO_4\\ Z:CuO\\ Các.PTHH:CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\\ Cu\left(NO_3\right)_2\rightarrow\left(t^o\right)CuO+2NO_2+\dfrac{1}{2}O_2\\ CuO+CO\rightarrow\left(t^o\right)Cu+CO_2\\ Cu+2H_2SO_{4\left(đặc\right)}\rightarrow\left(t^o\right)CuSO_4+SO_2+2H_2O\\ CuSO_4+BaCl_2\rightarrow BaSO_4+CuCl_2\)

Coi lại PTHH số 4

\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{Na_2SO_4}=b\left(mol\right)\left(a,b>0\right)\\ a,Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}106a+142b=24,8\\a=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ C_{MddHCl}=\dfrac{2.0,1}{0,2}=1\left(M\right)\\ b,\%m_{Na_2CO_3}=\dfrac{0,1.106}{24,8}.100\approx42,742\%\\ \Rightarrow\%m_{Na_2SO_4}\approx57,258\%\\ c,m_{muối}=m_{NaCl}+m_{Na_2SO_4}=2a.58,5+142b=25,9\left(g\right)\)

a) Ta có: \(m_{dd.tăng}=m_{KL}-m_{H_2}\Rightarrow m_{H_2}=11-10,2=0,8\left(g\right)\)

`=>` \(n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\Rightarrow56x+27y=11\left(1\right)\)

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

                 x---------------------------->x

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

                  y------------------------------->1,5y

`=> x + 1,5y = 0,4 (2)`

Từ `(1), (2) => x = 0,1; y = 0,2`

`=>` \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,9\%\\\%m_{Al}=100\%-50,9\%=49,1\%\end{matrix}\right.\)

b) \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

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

Áp dụng ĐLBTKL: \(m_{KClO_3\left(bđ\right)}=0,6.32+42,05=61,25\left(g\right)\)

`=>` \(n_{KClO_3\left(bđ\right)}=\dfrac{61,25}{122,5}=0,5\left(mol\right)\Rightarrow H=\dfrac{0,4}{0,5}.100\%=80\%\)

c) Ta có: \(\left\{{}\begin{matrix}m_C=m_X+m_Y=m_{H_2}+m_{O_2}=0,8+32.0,6=20\left(g\right)\\n_C=n_X+n_Y=n_{H_2}+n_{O_2}=0,4+0,6=1\left(mol\right)\end{matrix}\right.\)

`=>` \(M_C=\dfrac{20}{1}=20\left(g/mol\right)\)

`=>` \(d_{C/He}=\dfrac{20}{4}=5\)

\(a,n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{HCl}=1.V=V\left(mol\right);n_{H_2SO_4}=0,75.V\left(mol\right)\\ \Rightarrow n_{H\left(trong.axit\right)}=V+0,75.V.2=2,5V\left(mol\right)=2.n_{H_2}=2.n_{Mg}=2.0,15=0,3\left(mol\right)\\ \Leftrightarrow V=\dfrac{0,3}{2,5}=0,12\left(lít\right)\\ b,n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

\(\dfrac{C_{M_{HCl}}}{C_{M_{H_2SO_4}}}=\dfrac{1}{0.75}=\dfrac{4}{3}\)

\(n_{HCl}=a\left(mol\right)\Rightarrow n_{H_2SO_4}=\dfrac{3a}{4}\left(mol\right)\)

\(n_{Mg}=\dfrac{3.6}{24}=0.15\left(mol\right)\)

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

\(0.5a......a\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(\dfrac{3a}{4}.....\dfrac{3a}{4}\)

\(n_{Mg}=0.5a+\dfrac{3a}{4}=1.25a=0.15\left(mol\right)\)

\(\Rightarrow a=0.12\)

\(V=\dfrac{0.12}{1}=0.12\left(l\right)=120\left(ml\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(a,n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{HCl}=1.V=V\left(mol\right);n_{H_2SO_4}=0,75.V\left(mol\right)\\ \Rightarrow n_{H\left(trong.axit\right)}=V+0,75.V.2=2,5V\left(mol\right)=2.n_{H_2}=2.n_{Mg}=2.0,15=0,3\left(mol\right)\\ \Leftrightarrow V=\dfrac{0,3}{2,5}=0,12\left(lít\right)\\ b,n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

Em ơi bài này chưa cho nồng độ mol dung dịch HCl 

Em check lại đề giùm anh hi

a. \(n_{MOH}=0.25mol\)

Khí Y là \(H_2\Rightarrow n_{H_2}=\dfrac{0.6}{24}=0.025mol\)

\(2M+2H_2O\rightarrow2MOH+H_2\) (1)

0.05                    0.05   \(\leftarrow\) 0.025    (mol)

\(n_{MOH}\) tạo ra sau phản ứng (1) \(=0.25-0.05=0.2mol\)

\(M_2O+H_2O\rightarrow2MOH\)

 0.1                \(\leftarrow\)   0.2      (mol)

Ta có: \(0.05\times M_M+0.1\times\left(2M_M+16\right)=7.35\Leftrightarrow M_M=23\Rightarrow\) M là Natri

b.  \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

      0.25    \(\rightarrow\)    0.125         0.125

\(n_{H_2SO_4}=2\times0.15=0.3mol\)

Ta có: \(\dfrac{0.25}{2}< \dfrac{0.3}{1}\Rightarrow H_2SO_4\) dư

\(C_{M_{Na_2SO_4}}=\dfrac{0.125}{0.15}=0.83M\)

\(C_{M_{H_2SO_4}}\) dư \(=\dfrac{\left(0.3-0.125\right)}{0.15}=1.17M\)