\(n_{Cu\left(NO_3\right)_2}=\dfrac{64,86}{188}=0,345\left(mol\right)\)
Có: \(\left\{{}\begin{matrix}n_{N_2}+n_{H_2}=\dfrac{5,04}{22,4}=0,225\\\dfrac{28.n_{N_2}+2.n_{H_2}}{n_{N_2}+n_{H_2}}=11,4.2=22,8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{N_2}=0,18\left(mol\right)\\n_{H_2}=0,045\left(mol\right)\end{matrix}\right.\)
\(n_{NO_2}+n_{O_2}=\dfrac{6,272}{22,4}=0,28\left(mol\right)\)
- Nếu trong G không chứa muối amoni
Bảo toàn N: \(n_{NO_2}=0,345.2-0,18.2=0,33>0,28\) (vô lí)
=> Trong G chứa muối amoni
Bảo toàn O: \(6.n_{Cu\left(NO_3\right)_2}=n_{O\left(Y\right)}+2.n_{NO_2}+2.n_{O_2}\)
=> nO(Y) = 1,51 (mol)
=> \(n_{H_2O}=1,51\left(mol\right)\)
Bảo toàn H: \(1.n_{HCl}=2.n_{H_2}+2.n_{H_2O}+4.n_{NH_4Cl}\)
=> \(n_{NH_4Cl}=0,09\left(mol\right)\)
Bảo toàn Cu: \(n_{CuCl_2}=0,345\left(mol\right)\)
Bảo toàn Cl: \(1.n_{HCl}=2.n_{MgCl_2}+2.n_{CuCl_2}+1.n_{NH_4Cl}\)
=> \(n_{MgCl_2}=1,345\left(mol\right)\)
\(m=m_{MgCl_2}+m_{CuCl_2}+m_{NH_4Cl}=1,345.95+0,345.135+0,09.53,5=179,165\left(g\right)\)
=> D
