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)

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

\(a.n_{H_2}=\dfrac{14,874}{24,79}=0,6mol\\ \Rightarrow n_{H_2}=1,2mol\\ m_{HCl}=1,2.36,5=43,8g\\ b.m_{H_2}=0,6.2=1,2g\\ BTKL:m_X+m_{HCl}=m_{muối}+m_{H_2}\\ =>m_{muối}=m_X+m_{HCl}-m_{H_2}\\ =29,9+43,8-1,2\\ =72,5g\)

Giúp mình với ạ mình cần gấp ạ!!!

\(â,Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b,Đề có thiếu thể tích dung dịch sulfrue ko bạn?

Gọi: n_Mg = x (mol), n_Zn = 2x (mol)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Zn}=x+2x=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

⇒ x = 0,1 (mol)

⇒ n_Mg = 0,1 (mol), n_Zn = 0,2 (mol)

⇒ m_hh = 0,1.24 + 0,2.65 = 15,4 (g)

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

\(\Rightarrow V_{HCl}=\dfrac{0,6}{2}=0,3\left(l\right)\)

\(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

\(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

\(n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=0,1mol\\ m_{ZnCl_2}=0,1.136=13,6g\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

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

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)