a, Ta có nZn = \(\dfrac{16,25}{65}\) = 0,25 ( mol )
Zn + H2SO4 → ZnSO4 + H2
0,25\(\rightarrow\)0,25 \(\rightarrow\) 0,25 ----->0,25
=> VH2 = 22,4 . 0,25 = 5,6 ( lít )
Ta có Mdung dịch = Mtham gia - M\(\uparrow\)H2
= mZn + mH2SO4 - mH2
= 16,25 + 200 - 0,25 . 2
= 215,75
=> mH2SO4 = 98 . 0,25 = 24,5 ( gam )
=> C%H2SO4 = \(\dfrac{24,5}{215,75}\) . 100 \(\approx\) 11,4%
=> mZnSO4 = 161 . 0,25 = 40,25 ( gam )
=> C%ZnSO4 = \(\dfrac{40,25}{215,75}\) .100 \(\approx\) 18,7 %