Gọi: \(\left\{{}\begin{matrix}n_{N_2O_5}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\)

⇒ 108x + 44y = 27,2 (1)

Mà: \(d_{A/H_2}=34\Rightarrow M_A=34.2=68\left(g/mol\right)\)

\(\Rightarrow n_A=\dfrac{27,2}{68}=0,4\left(mol\right)=x+y\left(2\right)\) 

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,25\left(mol\right)\end{matrix}\right.\)

BTNT O, có: n_{O (trong A)} = 5n_N2O5 + 2n_CO2 = 1,25 (mol)

Ta có: n_{O (trong Fe2(SO4)3)} = 12n_Fe2(SO4)3

⇒ 12n_Fe2(SO4)3 = 1,25

⇒ n_Fe2(SO4)3 = 5/48 (mol)

⇒ m_Fe2(SO4)3 = 5/48.400 = 125/3 (g)

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Gọi KLK cần tìm là A.

PT: \(2A+2H_2O\rightarrow2AOH+H_2\)

Theo PT: \(n_A=2n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{7,8}{0,2}=39\left(g/mol\right)\)

Vậy: Kim loại kiềm đó là Kali.

Bạn tự làm phần tính chất hóa học nhé.

$\rm b)\\ CH_4+Cl_2\xrightarrow{askt}CH_3Cl+HCl\\c)\\ C_2H_4+Br_2\to C_2H_4Br_2\\C_2H_2+2Br_2\xrightarrow{1:2}C_2H_2Br_4\\d)\\CH_3COOH+C_2H_5OH\overset{H_2SO_4\ (đặc),\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O\\CH_3COOH+Na\to CH_3COONa+0,5H_2\\e)\\CH_3COOH+K\to CH_3COOK+0,5H_2\\CH_3COOH+KOH\to CH_3COOK+H_2O\\CH_3COOH+K_2O\to 2CH_3COOK+H_2O\\2CH_3COOH+K_2CO_3\to 2CH_3COOK+CO_2+H_2O$

Số ống nghiệm xảy ra phản ứng là 5: HCl,H2SO4 đặc nóng, AgNO3, ZnCl2,FeCl3

\(n_{O_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right);n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ 2Zn+O_2\rightarrow\left(t^o\right)2ZnO\\ Vì:\dfrac{0,1}{2}< \dfrac{0,2}{1}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,2-\dfrac{1}{2}.0,1=0,15\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\\ b,n_{ZnO}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnO}=81.0,1=8,1\left(g\right)\)

`#\text{lhg}`

`a)`

Ta có:

\(\text{n}_{\text{Zn}}=\dfrac{6,5}{65}=0,1\left(\text{mol}\right)\)

\(\text{n}_{\text{O}_2}=\dfrac{4,958}{24,79}=0,2\left(\text{mol}\right)\)

PTHH: \(\text{2Zn}+\text{O}_2\rightarrow\text{ 2ZnO}\)

Lập tỉ số: \(\dfrac{0,1}{2}:\dfrac{0,2}{1}=0,05< 0,2\)

Vậy, sau phản ứng Zn hết, O₂ dư

\(\text{n}_{\text{O}_2\text{ dư}}=0,2-\dfrac{0,1}{2}=0,15\left(\text{mol}\right)\)

\(\text{m}_{\text{O}_2\text{ dư}}=0,15\cdot32=4,8\left(\text{g}\right).\)

`b)`

Theo PT: \(\text{n}_{\text{ZnO}}=\text{n}_{\text{Zn}}\Rightarrow\text{n}_{\text{ZnO}}=0,1\left(\text{mol}\right)\)

Khối lượng oxide tạo thành:

\(\text{m}_{\text{ZnO}}=n_{\text{ZnO}}\cdot\text{M}_{\text{ZnO}}=0,1\cdot\left(65+16\right)=8,1\left(\text{g}\right).\)

tao

\(2Fe+3Cl_2\xrightarrow[]{t^0}2FeCl_3\)

Nguyên tắc là đổ từ từ axit vào nước em nhé!

đổ cả 2 cùng lúc:D

\(n_{H_2\left(đkc\right)}=\dfrac{11,1555}{24,79}=0,45\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Đặt:n_{Fe}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+1,5b=0,45\\56a+27b=19,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\left(TM\right)\\b=0,1\left(TM\right)\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,1.27}{19,5}.100\%\approx13,846\%\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2\left(tổng\right)}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ Đặt:a=n_{Fe};b=n_{Al}\left(a,b>0\right)\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}56a+27b=8,3\\a+1,5b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(TM\right)\\b=0,1\left(TM\right)\end{matrix}\right.\\ \%m_{Fe}=\dfrac{0,1.56}{8,3}.100\%\approx67,47\%\Rightarrow\%m_{Al}\approx32,53\%\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}m_{hh}=56x+27y=8,3\\n_{H_2}=x+\dfrac{3}{2}y=0,25\end{matrix}\right.\\ x=-2,6;y=5,5\)

Em xem lại đề giúp chị nhé!

ra chứ em, x=0,1; y=0,1 

Em ráp lại vào hpt của em hí