a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_{2\downarrow}+Na_2SO_4\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

b, \(m_{CuSO_4}=160.10\%=16\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{CuSO_4}=0,2\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

c, \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)

Ta có: \(n_{Cu}=\dfrac{8,32}{64}=0,13\left(mol\right)\)

\(n_{O_2}=\dfrac{6,81725}{24,79}=0,275\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(2Cu+O_2\underrightarrow{t^o}2CuO\)

Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}+\dfrac{1}{2}n_{Cu}\) \(\Rightarrow n_{Fe}=0,315\left(mol\right)\)

\(\Rightarrow a=0,315.56=17,64\left(g\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,105\left(mol\right)\\n_{CuO}=n_{Cu}=0,13\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%n_{Fe_3O_4}=\dfrac{0,105}{0,105+0,13}.100\%\approx44,7\%\\\%n_{CuO}\approx55,3\%\end{matrix}\right.\)

\(\overline{M_{hh}}=\dfrac{m_{Fe_3O_4}+n_{CuO}}{n_{Fe_3O_4}+n_{CuO}}=\dfrac{0,105.232+0,13.80}{0,105+0,13}=147,91\left(g/mol\right)\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Cu}+\dfrac{3}{4}n_{Al}\)

⇒ n_Cu = 0,1 (mol) 

⇒ m = 0,1.64 = 6,4 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{6,4}{6,4+5,4}.100\%\approx54,2\%\\\%m_{Al}\approx45,8\%\end{matrix}\right.\)

a, Ta có: 23n_Na + 39n_K = 2,94 (1)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{1,12}{22,4}=0,05\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,06\left(mol\right)\\n_K=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,06.23}{2,94}.100\%\approx46,9\%\\\%m_K\approx53,1\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,06\left(mol\right)\\n_{KOH}=n_K=0,04\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 2,94 + 97,16 - 0,05.2 = 100 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{0,06.40}{100}.100\%=2,4\%\\C\%_{KOH}=\dfrac{0,04.56}{100}.100\%=2,24\%\end{matrix}\right.\)

Ta có: n_FeSO4 = n_CuSO4 = x (mol)

- Khi nhúng M vào FeSO₄:

\(2M+nFeSO_4\rightarrow M_2\left(SO_4\right)_n+nFe\)

2x/n_______x_________________x (mol)

⇒ m _tăng = m_Fe - m_M = 56x - 2x.M_M/n = 16 (1)

- Khi nhúng vào CuSO₄:

\(2M+nCuSO_4\rightarrow M_2\left(SO_4\right)_n+nCu\)

2x/n________x_________________x (mol)

⇒ m _tăng = m_Cu - m_M = 64x - 2x.M_M/n = 20 (2)

Từ (1) và (2) ⇒ x = 0,5 (mol)

⇒ M_M = 12n

Với n = 2 thì M_M = 24 (g/mol)

→ M là Mg.

a, n_H⁺ = n_HCl + 2n_H2SO4 = 0,4.1 + 2.0,4.2 = 2 (mol)

Giả sử hh chỉ gồm Mg.

\(\Rightarrow n_{Mg}=\dfrac{12,9}{24}=0,5375\left(mol\right)\)

Xét: \(Mg+2H^+\rightarrow Mg^{2+}+H_2\)

có \(\dfrac{0,5375}{1}< \dfrac{2}{2}\) ta được H⁺ dư, mà n_hh max → dd C còn acid dư.

b, Gọi: \(\left\{{}\begin{matrix}n_{Mg}=3x\left(mol\right)\\n_{Fe}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\) ⇒ 3x.24 + 56x + 65y = 21,9 (1)

Có: \(n_{H_2}=n_{Mg}+n_{Fe}+n_{Zn}=3x+x+y=\dfrac{7,437}{24,79}=0,3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,05\left(mol\right)\\n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{12,9}.100\%\approx27,9\%\\\%m_{Fe}=\dfrac{0,05.56}{12,9}.100\%\approx21,7\%\\\%m_{Zn}\approx50,4\%\end{matrix}\right.\)

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b, \(n_{CH_4}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

\(n_{O_2}=2n_{CH_4}=1,2\left(mol\right)\Rightarrow V_{O_2}=1,2.24,79=29,748\left(l\right)\)

c, \(n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\Rightarrow m_{CO_2}=0,6.44=26,4\left(g\right)\)

d, \(V_{kk}=5V_{O_2}=148,92\left(l\right)\)

e, \(d_{CH_4/kk}=\dfrac{16}{29}\approx0,55< 1\)

→ CH₄ nhẹ hơn không khí, bằng 0,55 lần không khí.

a, Giả sử hỗn hợp chỉ gồm Zn.

Ta có: \(n_{Zn}=\dfrac{21}{65}\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{\dfrac{21}{65}}{1}>\dfrac{0,6}{2}\), ta được KL dư, mà n_hh min → A không tan hết.

b, Sửa đề: Cu → CuO

 \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

\(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,3}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,3\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,2\left(mol\right)\)

⇒ m _rắn = m_Cu + m_CuO = 0,3.64 + 0,2.80 = 35,2 (g)

H₂ không khử được MgO em nhé.

Do NaOH dư nên Zn(OH)₂ cũng bị hòa tan em nhé.