Giả sử a = 100g => mH2 = 5(g)
PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)(1)
PTHH:\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)(2)
\(PTHH:2Na+H_2O\rightarrow NaOH+H_2\)(3)
PTHH: \(NaOH+Mg\left(SO_4\right)\rightarrow Mg\left(OH\right)_2+Na_2SO_4\)(Pu này ko liên quan đến đề bài nhưng nó vẫn sảy ra nên mk vt).
Theo (1);(2), \(n_{H_2SO_4}=n_{H_2}=a\)
Theo (3), \(n_{H_2O}=2n_{H_2}=b\)
Ta có HPT: \(\left\{{}\begin{matrix}a+b=\dfrac{4,5}{2}\\98.a+18.2.b=100\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{19}{62}\\b=\dfrac{241}{124}\end{matrix}\right.\)
=>\(m_{H_2SO_4}=\dfrac{19}{62}.98=30,03\left(g\right)\)
=> \(A\%=\dfrac{30,03}{100}.100\%=30,03\%\)
