Giải thích các bước giải:
a. Gọi kim loại là M, hoá trị x
\(M+2xHNO_3\to M\left(NO_3\right)_x+xNO_2+xH_2O\)
\(2M+2xH_2SO_4\to M_2\left(SO_4\right)_x+xSO_2+2xH_2O\)
\(n_A=\dfrac{4,2}{22,4}=0,1875\ (mol)\)
\(\xrightarrow{\text{Theo PTHH}} n_{NO_3^-}+n_{SO_4^{2-}}=n_A=0,1875\ (mol)\)
Khi \(n_{NO_3^-}=0\ (mol)\)
\(\to n_{SO_4^{2-}}=0,1875\ (mol)\)
\(\to m_{\text{muối}}=6+0,1875.96=24\ (gam)\)
Khi \(n_{SO_4^{2-}}=0\ (mol)\)
\(\to n_{NO_3^-}=0,1875\ (mol)\)
\(\to m_{\text{muối}}=6+0,1875.62=17,625\ (gam)\)
\(\to 17,625<m_{\text{muối}}<24\)
b. Khi đó, ta có: \(46n_{NO_2}+64n_{SO_2}=9,75\ (gam); n_{NO_2}+n_{SO_2}=0,1875\ (mol)\)
\(\to n_{NO_2}=0,125\ (mol); n_{SO_2}=0,0625\ (mol)\)
\(\to m=6+0,125\cdot 62+0,0625\cdot 96=19,75\ (gam)\)