\(a)n_{Al_2O_3}=\dfrac{102}{102}=1kmol\\ n_{Al\left(lt\right)}=1\cdot\dfrac{4}{2}=2kmol\\ m_{Al\left(lt\right)}=2.27=54kg\\ H=\dfrac{51,3}{54}\cdot100\%=95\%\\ b)n_{Al}=\dfrac{54}{27}=2kmol\\ n_{Al_2O_3\left(lt\right)}=2\cdot\dfrac{2}{4}=1kmol\\ m_{Al_2O_3\left(lt\right)}=1.102=102kg\\ m_{Al_2O_3\left(tt\right)}=\dfrac{102}{92\%}\cdot100\%\approx110,87kg\)

Anh ơi sao H = 92% mà đoạn \(m_{Al_2O_3\left(tt\right)}\) lại là 95% vậy ạ?

xem lại đề !

\(n_{H_2}=\dfrac{1,456}{22,4}=0,065mol\\ BTNT\left(H\right):n_{HCl}=2n_{H_2}\\ \Leftrightarrow n_{HCl}=2.0,065=0,13mol\\ BTKL:m_{kl}+m_{HCl}=m_{muối}+m_{khí}\\ \Leftrightarrow1,61+0,13.36,5=m_{muối}+0,065.44\\ \Leftrightarrow m_{muối}=3,495g\)

anh Nhân đỉnh dữ, còn nhớ tới Hóa luôn :(

- Trích mẫu thử.

- Nhỏ từng mẫu thử vào giấy quỳ tím.

+ Quỳ hóa xanh: Ca(OH)₂

+ Quỳ hóa đỏ: H₂SO₄

+ Quỳ không đổi màu: Na₂SO₄

- Dán nhãn.

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)

a)

\(m_{Na_2CO_3}=n.M=0,2.106=21,2\left(g\right)\)

b)

Đổi:\(200ml=0,2l\)

\(n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

c)

\(m_{HCl}=\dfrac{300.14,6}{100}=43,8\left(g\right)\)

a, m_Na2CO3 = 0,2.106 = 21,2 (g)

b, n_NaOH = 0,2.1,5 = 0,3 (mol)

⇒ m_NaOH = 0,3.40 = 12 (g)

c, m_HCl = 300.14,6% = 43,8 (g)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\\ a)Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\\ Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ b)n_{H_2SO_4}=n_{CaCO_3}=n_{CO_2}=0,2mol\\ m_{H_2SO_4}=0,2.98=19,6g\\ C_{\%H_2SO_4}=\dfrac{19,6}{200}\cdot100\%=9,8\%\\ c)m_{CaCO_3}=0,2.100=20g\)

Bạn bổ sung đầy đủ đề nhé.

Ta có: 84n_MgCO3 + 100n_CaCO3 = 18,4 (1)

PT: \(MgCO_3+H_2SO_4\rightarrow MgSO_4+CO_2+H_2O\)

\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\)

Theo PT: \(n_{CO_2}=n_{MgCO_3}+n_{CaCO_3}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ n_MgCO3 = n_CaCO3 = 0,1 (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{MgSO_4}=n_{MgCO_3}=0,1\left(mol\right)\\n_{CaSO_4}=n_{CaCO_3}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow C_{M_{MgSO_4}}=C_{M_{CaSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\\ n_{MgCO_3}=a;n_{CaCO_3}=b\\ MgCO_3+H_2SO_4\rightarrow MgSO_4+CO_2+H_2O\)

a                   a                a                a             a

\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\)

b                   b                b                b             b

 \(\Rightarrow\left\{{}\begin{matrix}84a+100b=18,4\\a+b=0,2\end{matrix}\right.\\ \Rightarrow a=b=0,1mol\\ C_{M_{MgSO_4}}=\dfrac{0,1}{0,2}=0,5M\\ C_{M_{CaSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10,5}.100\%\approx61,9\%\\\%m_{Cu}\approx38,1\%\end{matrix}\right.\)

c, \(n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{Zn}=n_{H_2SO_4}=n_{H_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ \%m_{Zn}=\dfrac{6,5}{10,5}\cdot100\%=61,9\%\\ \%m_{Cu}=100\%-61,9=38,1\%\\ c.C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)