\(a,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,500ml=0,5l\\ n_{H_2SO_4}=0,5.1=0,5mol\\ n_{Zn}=\dfrac{13}{65}=0,2mol;n_{Fe}=\dfrac{14}{56}=0,25mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0,2.....0,2...........0,2..........0,2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0,25...0,25..........0,25......0,25\)
\(\Rightarrow n_{H_2SO_4.pứ}< n_{H_2SO_4}\left(0,2+0,25< 0,5\right)\\ \Rightarrow H_2SO_4.dư\\ V_{H_2}=\left(0,2+0,25\right)22,4=10,08l\\ c,m_{dd.lúc.đầu}=13+14+500.1,14=597g\\ m_{nước.thêm.vào}=600-597=3g\\ C_{\%H_2SO_4.dư}=\dfrac{\left(0,5-0,2-0,25\right).98}{600}\cdot100\%=0,82\%\\ C_{\%ZnSO_4}=\dfrac{0,2.161}{600}\cdot100\%=5,37\%\\ C_{\%FeSO_4}=\dfrac{0,25.152}{600}\cdot100\%=6,33\%\)
