Chẳng ai giải đc

theo đề ra ta có 

dx_/kok = \(\dfrac{M_X}{29}\) = 0,3276 => Mx =9,5

gọi x và y lần lượt là số mol của H₂ và O₂ 

n_x = x + y

m_x = 2x + 32y

\(\overline{M}\) = \(\dfrac{2x+32y}{x+y}=9,2\)

x = 22,5    y =77,5

Giả sử có 1 mol khí Cl₂, 2 mol khí O₂

a) \(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{1}{1+2}.100\%=33,33\%\\\%V_{O_2}=\dfrac{2}{1+2}.100\%=66,67\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Cl_2}=\dfrac{1.71}{1.71+2.32}.100\%=52,59\%\\\%m_{O_2}=\dfrac{2.32}{1.71+2.32}.100\%=47,41\%\end{matrix}\right.\)

b) \(\overline{M}=\dfrac{1.71+2.32}{1+2}=45\left(g/mol\right)\)

=> \(d_{A/H_2}=\dfrac{45}{2}=22,5\)

c) \(n_A=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> m_A = 0,3.45 = 13,5 (g)

a) \(\overline{M}_A=5,875.2=11,75\left(g/mol\right)\)

b) Gọi số mol N₂, H₂ là a, b (mol)

\(\overline{M}_A=\dfrac{28a+2b}{a+b}=11,75\left(g/mol\right)\)

=> 16,25a = 9,75b 

=> a = 0,6b 

\(\left\{{}\begin{matrix}\%n_{N_2}=\dfrac{a}{a+b}.100\%=\dfrac{0,6b}{0,6b+b}.100\%=37,5\%\\\%n_{H_2}=\dfrac{b}{a+b}.100\%=\dfrac{b}{0,6b+b}.100\%=62,5\%\end{matrix}\right.\)

c) 

1 mol hỗn hợp A chứa \(\left\{{}\begin{matrix}n_{N_2}=\dfrac{1.37,5}{100}=0,375\left(mol\right)\\n_{H_2}=\dfrac{1.62,5}{100}=0,625\left(mol\right)\end{matrix}\right.\)

\(\overline{M}_B=\dfrac{0,375.28+0,625.2+17x}{1+x}=6,4.2=12,8\left(g/mol\right)\)

=> x = 0,25 (mol)

\(a.\)

\(GS:\)

\(n_{hh}=1\left(mol\right)\)

\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)

\(\Rightarrow a+b=1\left(1\right)\)

\(m_A=28a+44b=18\cdot2=36\left(2\right)\)

\(\left(1\right),\left(2\right):\)

\(a=b=0.5\)

\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)

\(\%m_{CO_2}=61.11\%\)

\(b.\)

\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)

\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)

\(Đặt:n_{CO_2}=x\left(mol\right)\)

\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow x=0.2\)

\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)

Vì Mg có tính khử rất mạnh nên trong Y có thể có muối (NH₄)₂SO₄

BTNT N: 2n_Fe(NO3)2 = n_NO + n_NH4 => 2y = 0,02 + n_NH4 => n_NH4 = 2y – 0,02

BTNT H: 2n_H2SO4 = 2n_H2 + 4n_NH4 + 2n_H2O => 2 . 0,39 = 2 . 0,05 + 4 (2y – 0,02) + 2n_H2O

=> n_H2O = 0,38 – 4y

BTNT O: n_FeO + 6n_Fe(NO3)2 = n_NO + n_H2O

=> x + 6y = 0,02 + 0,38 – 4y => x + 10y = 0,4

Gọi \(n_{H_2} = a(mol) ; n_{N_2} = b(mol)\)

Coi \(n_{hỗn\ hợp} = 1(mol)\)

Ta có :

\(n_{hỗn\ hợp} = a + b = 1(mol)\\ m_{hỗn\ hợp} = 2a + 28b = 21,5.1 = 21,5(gam)\\ \Rightarrow a = 0,25 ; b = 0,75\)

Vậy :

\(\%V_{H_2} = \dfrac{0,25}{1}.100\% = 25\%\\ \%V_{N_2} = 100\% - 25\% = 75\%\)