Question

tìm C% dd thu đc

150g dd K₂SO₃ 7,6% vào 250g dd H₂SO₄ 9,8%

Answer 1

K2SO3 + H2SO4

Answer 2

PTHH: K2SO3 + H2SO4 -> H2O + SO2 + K2SO4
mK2SO3=150*7,6%=11,4g<->0,07(mol)
mH2SO4=250*9,8%=24,5g<->0,25(mol)
-Theo pt: nK2SO3=nH2SO4=0,07 mol < 0,25 mol
=>nH2SO4 dư=0,25-0,07=0,18(mol)<->17,64g
-Theo pt: nK2SO4=nK2SO3=0,07(mol)<->12,18g
nSO2 = nK2SO3 = 0,07 mol -> 4,48g

nH2O = nK2SO3 = 0,07 mol-> 1,26g

-m dd thu đc=11,4+ 24,5+ 1,26 - 4,48=32,68g
C% H2SO4 dư=17,64/250 *100=7%
C% K2SO4=12,18/32,68 * 100=37,27%