Bài 1:
mdd=D.Vdd=200.1,05=210 g
m ct H2SO4=$\frac{210.9,8}{100}$ =20,58 g
n H2SO4=$\frac{20,58}{98}$ =0,21 mol
2NaOH+H2SO4→Na2SO4+H2O
0,42 ← 0,21 → 0,21 mol
m ct NaOH=0,42.40=16,8 g
a.mdd NaOH=$\frac{16,8.100}{10}$ =168 g
m ct Na2SO4=0,21.142=29,82 g
mdd sau=mdd H2SO4+mdd NaOH=168+210=378 g
b.C% Na2SO4=$\frac{29,82.100}{378}$ ≈7,889 %
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