a, Bạn bổ sung đề phần này nhé.

b,

1. \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

2. \(2Ca+O_2\underrightarrow{t^o}2CaO\)

3. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

4. \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

- Trích mẫu thử.

- Nhỏ vài giọt từng mẫu thử vào giấy quỳ tím.

+ Quỳ hóa đỏ: HCl

+ Quỳ hóa xanh: NaOH

+ Quỳ không đổi màu: NaCl

- Dán nhãn.

a, \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

b, Ta có: \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,2.100=20\left(g\right)\)

c, \(n_{\left(CH_3COO\right)_2Ca}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Ca}=0,1.158=15,8\left(g\right)\)

Đề yêu cầu gì bạn nhỉ?

a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

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

c, \(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)

d, \(CaCl_2+Na_2CO_3\rightarrow CaCO_3+2NaCl\)

e, \(3Fe_3O_4+8Al\underrightarrow{t^o}4Al_2O_3+9Fe\)

g, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____0,1_____0,2______0,1____0,1 (mol)

a, V_H2 = 0,1.24,79 = 2,479 (l)

b, m_MgCl2 = 0,1.95 = 9,5 (g)

c, \(C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

a, \(CH_3-\left[CH_2\right]_5-CH_3\underrightarrow{^{t^o,xt,p}}C_6H_5CH_3+4H_2\)

\(C_5H_6CH_3+3HNO_3\underrightarrow{^{H_2SO_{4đ},}}C_6H_2CH_3\left(NO_2\right)_3+3H_2O\)

b, Ta có: 0,4 tấn = 400 (kg)

\(\Rightarrow n_{C_7H_{16}\left(LT\right)}=n_{C_6H_5CH_3\left(LT\right)}=n_{C_6H_2CH_3\left(NO_2\right)_3}=\dfrac{400}{227}=\left(kmol\right)\)

\(\Rightarrow n_{C_7H_{16}\left(TT\right)}=\dfrac{400}{227}:70\%:40\%=\dfrac{10000}{1589}\left(kmol\right)\)

\(\Rightarrow m_{C_7H_{16}}=\dfrac{10000}{1589}.100=629,3\left(kg\right)\)

a, PT: \(Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+H_2O\)

\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+H_2O\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: n_HCl = 0,1.1 = 0,1 (mol)

Theo PT: \(n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

Theo ĐLBTKL, có: m + m_HCl = m _muối + m_H2O

⇒ m = 24,1 + 0,05.18 - 0,1.36,5 = 21,35 (g)

b, Phần không tan là Mg(OH)₂ và Cu(OH)₂

PT: \(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

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

\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Chất rắn là Cu.

\(\Rightarrow n_{Cu\left(OH\right)_2}=n_{CuO}=n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)=n_{CuCl_2}\)

\(n_{Mg\left(OH\right)_2}=n_{MgO}=n_{MgSO_4}=n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow n_{NaOH}=n_{NaCl}=\dfrac{24,1-m_{MgCl_2}-m_{CuCl_2}}{58,5}=\dfrac{157}{1170}\left(mol\right)\)

⇒ m_Mg(OH)2 + m_Cu(OH)2 + m_NaOH = 16,07 (g)

⇒ m _{tạp chất} = 21,35 - 16,07 = 5,28 (g)

\(n_{CO_2}=\dfrac{5,9496}{24,79}=0,24\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,24\left(mol\right)\)

a, \(V_{Ca\left(OH\right)_2}=\dfrac{0,24}{1}=0,24\left(l\right)\)

b, m_CaCO3 = 0,24.100 = 24 (g)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b, \(n_{C_2H_6O}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=2n_{C_2H_6O}=0,4\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)