Đáp án:
a)ZnO+ H2SO4 →ZnSO4 + H2O 10%=$\frac{m H2SO4}{250}$ . 100%=>mH2SO4=25
nđ 0,1.......$\frac{25}{98}$ ..............0.................0 =>nH2SO4=$\frac{25}{98}$
npư 0,1......0,1................0.................0
ns 0..........$\frac{38}{245}$ ......0,1......0,1
=>mH2SO4 dư=$\frac{38}{245}$.98=15,2
=>nZnSO4=0,1 nH2O=0,1
b)BTKL : mZnO + mddH2SO4 = mdd(sau)
8,1+ 250=258,1
C%=$\frac{0,1.161}{258,1}$ .100%=$\frac{16100}{2581}$ %≈6,24%