\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

Ta có: \(n_{FeO}=\dfrac{11,2}{72}=\dfrac{7}{45}\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{FeO}=\dfrac{7}{45}\left(mol\right)\)

\(\Rightarrow m_{FeSO_4}=\dfrac{7}{45}.152=23,64\left(g\right)\)

\(V_{ddH_2SO_4}=\dfrac{\dfrac{7}{45}}{0,1}=\dfrac{14}{9}\left(l\right)\)

\(\Rightarrow C_{M_{FeSO_4}}=\dfrac{\dfrac{7}{45}}{\dfrac{14}{9}}=0,1\left(M\right)\)

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca++CO_2+H_2O\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,1.60}{10,6}.100\%\approx56,6\%\\\%m_{C_2H_5OH}\approx43,4\%\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\)

Theo PT: n_CH3COOH = 2n_CO2 = 0,1 (mol)

Đến đây tính tiếp giống bài với CaCO₃ mình vừa trả lời bạn nhé.

a. \(Fe+CuSO_4\rightarrow FeSO_4+Cu\downarrow\)

b. \(CaCO_3+HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

c. \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

d. \(CuSO_4+BaCl_2\rightarrow CuCl_2+BaSO_4\downarrow\)

\(C_6H_{12}O_6\underrightarrow{^{t^o,xt}}2C_2H_5OH+2CO_2\)

Ta có: \(n_{C_6H_{12}O_6}=\dfrac{600}{180}=\dfrac{10}{3}\left(mol\right)\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=2n_{C_6H_{12}O_6}=\dfrac{20}{3}\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(LT\right)}=\dfrac{20}{3}.46=\dfrac{920}{3}\left(g\right)< m_{C_2H_5OH\left(TT\right)}\) → vô lý

Bạn xem lại đề nhé.

Câu 20. Đáp án đúng

B. BaO, CO₂, CuO, ZnO.

 X là Strontium( Sr) nha