PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\\n_{Ca\left(OH\right)_2}=\dfrac{150\cdot7,4\%}{74}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối

PTHH: \(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)

                        a_____2a_____________a                  (mol)

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

                       b_____b_________b                             (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}2a+b=0,2\\a+b=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)

\(\Rightarrow C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,05\cdot162}{0,2\cdot44+150-0,1\cdot100}\cdot100\%\approx5,44\%\)

a) Gọi \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow64a+27b=1,23\)  (1)

Bảo toàn electron: \(2n_{Cu}+3n_A=n_{NO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(\Rightarrow2a+3b=0,06\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,015\\b=0,01\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,015\cdot64}{1,23}\cdot100\%\approx78,05\%\\\%m_{Al}=21,95\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{CuO}=n_{Cu}=0,015\left(mol\right)\) \(\Rightarrow m_{CuO}=0,015\cdot80=1,2\left(g\right)\)

pt số 8 ko xảy ra nhé e :((

PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3\cdot98}{24+200}\cdot100\%=13,125\%\)

Các p.ứ oxi hóa - khử: b, c, e, f, g và h

a) Na₂SO₄ và Fe(OH)₂

b) Fe₂O₃

c) CH₄

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

b+c) Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)=n_{FeCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

d) Số phân tử H₂: \(0,2\cdot6\cdot10^{23}=1,2\cdot10^{23}\left(phân.tử\right)\)

e) 

+) Cách 1: Theo PTHH: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)

+) Cách 2:

Ta có: \(m_{H_2}=0,2\cdot2=0,4\left(g\right)\) 

Bảo toàn khối lượng: \(m_{HCl}=m_{FeCl_2}+m_{H_2}-m_{Fe}=14,6\left(g\right)\)

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

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{MgCl_2}=0,2\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,4\cdot36,5}{100}\cdot100\%=14,6\%\\m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{4,8+100-0,4}\cdot100\%\approx18,2\%\)