Đáp án D

Định hướng tư duy giải

Có; \(n_{NO}+n_{N_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\dfrac{30n_{NO}+28n_{N_2}}{n_{NO}+n_{N_2}}=\dfrac{43}{3}.2=\dfrac{86}{3}\)

=> n_NO = 0,1 (mol); n_N2 = 0,2(mol)

Al⁰ -3e --> Al⁺³

a-->3a--->a

Mg⁰ -2e --> Mg⁺²

b---->2b---->b

Fe⁰ -3e --> Fe⁺³

c--->3c---->c

4H⁺ + NO₃^- +3e --> NO + 2H₂O

0,4<----------0,3<----0,1

12H⁺ + 2NO₃^- +10e--> N₂ + 6H₂O

2,4<---------------2<-----0,2

10H⁺ + 2NO₃^- +8e--> NH₄NO₃ + 3H₂O

0,5<-------------0,4<-----0,05

=> n_HNO3 = n_H⁺ = 0,4 + 2,4 + 0,5 = 3,3 (mol)

=> \(V_{dd}=\dfrac{3,3}{1}=3,3\left(l\right)\)

Bảo toàn H: \(n_{H\left(HNO_3\right)}=n_{H\left(NH_4NO_3\right)}+n_{H\left(H_2O\right)}\)

=> 3,3 = 4.0,05 + 2.n_H2O

=> n_H2O = 1,55(mol)

Theo ĐLBTKL:

\(m_{hh\left(bd\right)}+m_{HNO_3}=m_{muoi}+m_{NH_4NO_3}+m_{NO}+m_{N_2}+m_{H_2O}\)

=> hh muỗi khan = \(m_{muoi}+m_{NH_4NO_3}=11,9+3,3.63-30.0,1-28.0,2-18.1,55=183,3\left(g\right)\)

=> A

Đặt \(n_{NO}=a(mol);n_{N_2}=b(mol)\)

\(\Rightarrow \begin{cases} a+b=\dfrac{6,72}{22,4}=0,3(mol)\\ \dfrac{30a+28b}{a+b}=2.\dfrac{43}{3}=\dfrac{86}{3} \end{cases}\Rightarrow \begin{cases} a=0,1(mol)\\ b=0,2(mol) \end{cases}\\ \Rightarrow \Sigma n_{HNO_3}=4n_{NO}+12n_{N_2}+10n_{NH_4NO_3}=3,3(mol)\\ \Rightarrow V_{dd_{HNO_3}}=\dfrac{3,3}{1}=3,3(l)\\ n_{NO_3-(muối)}=3n_{NO}+10n_{N_2}+8n_{NH_4NO_3}=2,7(mol)\\ \Rightarrow m_{muối}=m_{NO_3-(muối)}+m_{KL}+m_{NH_4NO_3}=2,7.62+11,9+0,05.80=183,3(g) \)

Chọn A

Ta có: n_N2 = 0,22 (mol)

⇒ n_NO3^- = 10n_N2 = 2,2 (mol)

⇒ m _muối = m_X + m_NO3^- = 31 + 2,2.62 = 167,4 (g)

n_HNO3 = 12n_N2 = 2,64 (mol)

\(\Rightarrow V_{HNO_3}=\dfrac{2,64}{2}=1,32\left(l\right)=1320\left(ml\right)\)

$n_{HCl} = 0,8.0,5 = 0,4(mol) ; n_{H_2SO_4} = 0,6(mol) ;n_{H_2} = 0,2(mol)$

$n_{H(trong\ axit)} = 0,4 + 0,6.2 = 1,6(mol)$

Bảo toàn H : $n_{H_2O} = \dfrac{n_{H(trong\ axit)} - 2n_{H_2} }{2} = 0,6(mol)$
Bảo toàn khối lượng : 

$m = 88,7 + 0,6.18 + 0,2.2 - 0,4.36,5 -  0,6.98 = 26,5(gam)$

\(n_{HCl}=0,8.0,5=0,4\left(mol\right);n_{H_2SO_4}=0,8.0,75=0,6\left(mol\right)\)

=> \(n_{Cl^-}=0,4\left(mol\right);n_{SO_4^{2-}}=0,6\left(mol\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Bảo toàn nguyên tố H:

\(n_{HCl}.1+n_{H_2SO_4}.2=n_{H_2}.2+n_{H_2O}.2\)

\(\Leftrightarrow0,4.1+0,6.2=0,2.2+n_{H_2O}.2\)

=>\(n_{H_2O}=0,6\left(mol\right)\)

=> \(n_O=0,6\left(mol\right)\)

\(m_{muối}=m_{kimloai}+m_{Cl^-}+m_{SO_4^{2-}}\)

=>\(m_{kimloai}=88,7-35,5.0,4-0,6.96=16,9\left(g\right)\)

=> \(m=m_{kimloai}+m_O=16,9+0,6.16=26,5\left(g\right)\)

n N2O = n N2 = 1,792/44,8 = 0,04 mol

--> m muối = 10,71 + 0,04 x 10 x 62 + 0,04 x 8 x 62 = 55,35 gam. 

n HNO3 = 0,04 x 12 + 0,04 x 10 = 0,88 mol

--> Cm HNO3 = 0,88/4 = 0,22 M. 

nN2=nN2O=0,04 mol⇒nNO3 muoi=10nN2+8nN2O=0,72 mol⇒mmuoi=mKL+mNO3 muoi=55,35 gnHNO3=4a=12nN2+10nN2O=0,88⇒a=0,22
M