SO3+H2O->H2SO4

m/80            49m/40   (mol)

mdd H2SO4 24,5%=500.1,2=600g

=>mH2SO4=600.24,5%=147g

m dd sau=m+600 (g)

m H2SO4 sau=147+1,225m (g)

C% sau=49% nên ta có:

$$\begin{gathered} \frac{147+1,225m}{m+600}=0,49<=>147+1,225m=0,49m+294 \\ <=>0,735m=147<=>m=200 \end{gathered}$$

=>mdd sau=800g; mH2SO4 sau=392g=>nH2SO4=4 mol

V sau=800/1,2=2000/3 ml

=>CM=6 M

mdung dịch H2SO4 = 500.1,2 =600(g)=

=> mH2SO4 = 600.24,5% =147(g)

PTHH : SO3 + H2O -> H2SO4

nSO3 = m/80(mol)

Theo PTHH => nH2SO4 = m/80 (mol) => mH2SO4 = 1,225m(g)

=> ΣmH2SO4=1,225m+147 (g)

mdung dịch = m + 600 (g)

Theo bài ra ta có hệ :

$\frac{1,225m+147}{m+600}.1005=49\%$=> m=200 (g)

m sau = 800g => nsau = 200/80 = 2,5

=> CM =2,5/0,5= 5M

SO3+H2O->H2SO4

m/80            49m/40   (mol)

mdd H2SO4 24,5%=500.1,2=600g

=>mH2SO4=600.24,5%=147g

m dd sau=m+600 (g)

m H2SO4 sau=147+1,225m (g)

C% sau=49% nên ta có:

147+1,225mm+600=0,49<=>147+1,225m=0,49m+294<=>0,735m=147<=>m=200147+1,225mm+600=0,49<=>147+1,225m=0,49m+294<=>0,735m=147<=>m=200

=>mdd sau=800g; mH2SO4 sau=392g=>nH2SO4=4 mol

V sau=800/1,2=2000/3 ml

=>CM=6 M

SO3+H2O->H2SO4

m/80            49m/40   (mol)

mdd H2SO4 24,5%=500.1,2=600g

=>mH2SO4=600.24,5%=147g

m dd sau=m+600 (g)

m H2SO4 sau=147+1,225m (g)

C% sau=49% nên ta có:

147+1,225mm+600=0,49<=>147+1,225m=0,49m+294<=>0,735m=147<=>m=200147+1,225mm+600=0,49<=>147+1,225m=0,49m+294<=>0,735m=147<=>m=200

=>mdd sau=800g; mH2SO4 sau=392g=>nH2SO4=4 mol

V sau=800/1,2=2000/3 ml

=>CM=6 M
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