Câu `3:`
`n_{SO_2}=\frac{12,88}{22,4}=0,575(mol)`
BTe:
$\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}$
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+3}+3e$
$\mathop{Ag}\limits^{0}\to \mathop{Ag}\limits^{+}+e$
$\mathop{Cu}\limits^{0}\to \mathop{Cu}\limits^{+2}+2e$
`∑n_{\text{e cho}}=∑n_{\text{e nhận}}=0,575.2=1,15(mol)`
Do `n_{H_2SO_4}=\frac{n_{\text{ e cho}}}{2}`
`=> n_{H_2SO_4}=0,575(mol)`
`m_{\text{muối}}=m_{\text{hh kim loại}}+m_{SO_4^{2-}}`
`=> m_{\text{muối}}=28,6+0,575.96=83,8g`
