\(V_{H_2}=10.0,2=2\left(l\right)\)
=> \(n_{H_2\left(thu.đc\right)}=\dfrac{2}{22,4}=\dfrac{5}{56}\left(mol\right)\) => \(n_{H_2\left(sinh.ra\right)}=\dfrac{5}{56}:85\%=\dfrac{25}{238}\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
\(\dfrac{25}{238}\)<--\(\dfrac{25}{119}\)<---------\(\dfrac{25}{238}\)
=> \(m_{Zn}=\dfrac{25}{238}.65=\dfrac{1625}{238}\left(g\right)\)
\(m_{HCl}=\dfrac{25}{119}.36,5=\dfrac{1825}{238}\left(g\right)\)