Giải

nO(B)=$\frac{28-20}{10}$=0,5=nH2O(BTO)

TH1:

BTH: 2nH2SO4pư=2nH2O

       <=>nH2SO4=nH20=0,5

       <=>nH2SO4=$\frac{20}{100}$nH2SO4pư

                            =0,5.$\frac{20}{100}$=0,1 

       <=>nH2SO4=0,6=>V=0.6(l)

   $m_{muối}$= $m_{khối lương}$+ $m_{SO4pứ}$ =20+0,5.96=68(g) 

TH2:
BTH=2nHCl+2nH2SO4pứ=nH2O

<=>2.Vpứ+Vpứ.2=0.5:2

<=>Vpứ=0,25(l)

nHClpứ=2.0p/25=0.5(l)

nH2SO4pứ=0.25(mol)

=>VddC=0.3(l)

$m_{muối}$= $m_{m}$+ $m_{SO4pứ}$+$m_{HClpứ}$ 

=20+0,5.35,5+0,25.96=61,25(g)

TH3:
BTH:$n_{H2SO4}$$^{-1}$ =$n_{H2O}$ $^{-2}$  

=>nNaOHpứ=2.45=1

=>nNaHSO4dư=$\frac{20}{100}$.1=0.2

=>nNaHSO4dư=0.5V

=>V=2,4(l)

$m_{muối}$= $m_{Napứ}$+ $m_{SO4}$+$m_{m}$

=1.23+1.96+0,2.120+20=163(g)

TH4:

BTH:nNaHSO4pứ+nH2SO4pứ=$n_{H2O}$ $^{-2}$ 

<=>0.5Vpứ+2.0,25Vpứ=0,5.2=>Vpứ=1(l)

=>nNaHSO4pứ=0.5+0.1=0.6=0.5V

=>Vbđ=1.21(l)

$m_{muối}$= $m_{m}$+ $m_{Napứ}$+$m_{SO4pứ}$+ $m_{NaHSO4pứ}$ 

=20+0,5.23+(0,5+0,25).96+0,1.120

=115,5(g)

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