=> m_N=n_N.14=0,3.14=4,2(g)

Nãy lỗi mạng quá

\(a.n_{N_2O}=\dfrac{8,8}{44}=0,2\left(mol\right)\\ \Rightarrow n_N=2n_{N_2O}=0,4\left(mol\right)\\ \Rightarrow m_N=0,4.14=5,6\left(g\right)\\ b.n_N=2n_{NH_4NO_3}=\dfrac{2.6.10^{23}}{6.10^{23}}=2\left(mol\right)\\ \Rightarrow m_N=2.14=28\left(g\right)\\ c.n_N=n_{NO}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_N=0,3.14=4,2\left(g\right)\)

\(\left\{{}\begin{matrix}n_{N_2O}+n_{NO}=0,04\\44n_{N_2O}+30n_{NO}=18,5.2.0,04=1,48\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,02\left(mol\right)\\n_{NO}=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{N_2O}=\dfrac{0,02.44}{1,48}.100\%=59,46\%\\\%m_{NO}=\dfrac{0,02.30}{1,48}.100\%=40,54\%\end{matrix}\right.\)

Gọi số mol N₂O, N₂ là a, b (mol)

Có: \(\overline{M}_A=\dfrac{44a+28b}{a+b}=18,45.2=36,9\left(g/mol\right)\)

=> 7,1a = 8,9b (1)

PTHH: 8Fe + 30HNO₃ --> 8Fe(NO₃)₃ + 3N₂O + 15H₂O

           \(\dfrac{8}{3}a\)<-------------------------------a

             10Fe + 36HNO₃ --> 10Fe(NO₃)₃ + 3N₂ + 18H₂O

              \(\dfrac{10}{3}b\)<------------------------------b

=> \(\dfrac{8}{3}a+\dfrac{10}{3}b=\dfrac{4,48}{22,4}=0,2\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{89}{2370}\left(mol\right)\\b=\dfrac{71}{2370}\left(mol\right)\end{matrix}\right.\) => \(V=\left(\dfrac{89}{2370}+\dfrac{71}{2370}\right).22,4=\dfrac{1792}{1185}\left(l\right)\)

a) \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_O=4n_{Fe_3O_4}=0,08\left(mol\right)\end{matrix}\right.\)

b) \(n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_N=2n_{N_2O}=0,3\left(mol\right)\\n_O=n_{N_2O}=0,15\left(mol\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_H=2n_{H_2SO_4}=0,1\left(mol\right)\\n_S=n_{H_2SO_4}=0,05\left(mol\right)\\n_O=4n_{H_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\)

Gọi số mol của N₂O, CO₂, C₃H₈ là a, b, c

=> \(\left\{{}\begin{matrix}a+b+c=\dfrac{3,36}{22,4}=0,15\\44a=44b=44c\end{matrix}\right.\)

=> a = b = c = 0,05 (mol)

Khí thoát ra gồm N₂O, C₃H₈

=> V = (0,05 + 0,05).22,4 = 2,24(l)

=> A

Gọi số mol của N₂O, CO₂, C₃H₈ là a, b, c

=> 

\(n_{CO_2}=a\left(mol\right),n_{N_2O}=b\left(mol\right),n_{H_2}=c\left(Mol\right)\)

\(n_A=a+b+c=0.05\left(mol\right)\)

\(\Leftrightarrow44a+44b+44c=2.2\left(1\right)\)

\(m_A=44a+44b+2c=1.78\left(g\right)\left(2\right)\)

\(\Rightarrow c=0.01\)

\(m_B=0.01\cdot2+0.03\cdot X=0.14\left(g\right)\)

\(\Rightarrow X=4\)

\(X:He\)