Đáp án:

$Zn(NO_3)_2$

Giải thích các bước giải:

$m_{cr\ giảm}=m_{NO_2}+m_{O_2}=8,64(g)\\ Muối:\ M(NO_3)_n\\ Oxit:\ M_2O_m\ (m\ge n)\\ \xrightarrow{\text{BTKL}}\ m_{M(NO_3)_n}=m_{oxit}+m_{NO_2}+m_{O_2}=15,12(g)\\ 4M(NO_3)_n\xrightarrow{t^0} 2M_2O_m+4nNO_2+(2n-m)O_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{M(NO_3)_n}=2.n_{M_2O_m}\\ \Rightarrow \dfrac{15,12}{M+62n}=2.\dfrac{6,48}{2M+16m}\\ \Rightarrow 17,28M+241,92m-803,52n=0\\ \Rightarrow 1728M+24192m-80352n=0\\ \Rightarrow \begin{cases}n=2\\m=2\\M=65\end{cases}\\ \to M:\ Zn\\ \to CTHH:\ Zn(NO_3)_2$

Đáp án:

`Zn(NO_3)_2`

Giải thích các bước giải:

 `CT\ m uoi: M(NO_3)_n`

`Pt: 4M(NO_3)_n` $\xrightarrow{t^o}$ `2M_2O_n+4nNO_2+nO_2`

`m_{g i am}=m_{NO_2}+m_{O_2}=8,64(g)`

`\text{BTKL:}\ m_{m uoi}=m_{o x i t}+m_{g iam}`

`<=>m_{m uoi}=6,48+8,64=15,12(g)`

`Ta\ co: n_{M(NO_3)_n}=2n_{M_2O_n}`

`<=>{15,12}/{M+62n}={2.6,48}/{2M+16n}`

`<=>30,24M+241,92n=12,96M+803,52n`

`<=>17,28M=561,6n`

`<=>M=32,5n`

`Lap\ bang:`

\begin{array}{|c|c|c|}\hline \text{n}&\text{1}&\text{2}&\text{3}\\\hline \text{M}&\text{32,5(loại)}&\text{65(chọn)}&\text{97,5(loại)}\\\hline\end{array}

`M uoi: Zn(NO_3)_2`