\(n_{H_2SO_4}=a\left(mol\right)\)
\(\Rightarrow n_{H_2}=a\left(mol\right)\)
\(BTKL:\)
\(17.5+98a=31.7+2a\)
\(\Rightarrow a=\dfrac{71}{480}\)
\(V_{dd_{H_2SO_4}}=\dfrac{\dfrac{71}{480}}{2}=0.074\left(l\right)\)
\(V_{H_2}=\dfrac{71}{480}\cdot22.4=3.313\left(l\right)\)
Ta có: \(m_{SO_4^{2-}}=m_{muối}-m_{KL}=14,2\left(g\right)\) \(\Rightarrow n_{SO_4^{2-}}=\dfrac{14,2}{32+16\cdot4}=\dfrac{71}{480}\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2SO_4}=\dfrac{\dfrac{71}{480}}{2}\approx0,074\left(l\right)\\V_{H_2}=\dfrac{71}{480}\cdot22,4\approx3,31\left(l\right)\end{matrix}\right.\)
