Đáp án:

$C_{M\,H_2SO_4(bđ)}=1,2M$ 

Giải thích các bước giải:

Đổi $250ml=0,25l;600ml=0,6l$

$n_{NaOH}=0,6.1=0,6(mol)$

$n_{BaCl_2}=\frac{500.10,4\%}{208}=0,25(mol)$

$BaCl_2+H_2SO_4\to BaSO_4↓+2HCl$

Theo PT: $n_{HCl}=2n_{BaCl_2}=0,5(mol);n_{H_2SO_4(pứ)}=n_{BaCl_2}=0,25(mol)$

$NaOH+HCl\to NaCl+H_2O$

$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

Theo PT: $2n_{H_2SO_4(dư)}+n_{HCl}=n_{NaOH}$

$\to 2n_{H_2SO_4(dư)}+0,5=0,6$

$\to n_{H_2SO_4(dư)}=0,05(mol)$

$\to \sum n_{H_2SO_4}=0,25+0,05=0,3(mol)$

$\to C_{M\,H_2SO_4(bđ)}=\frac{0,3}{0,25}=1,2M$

BaCl2 +H2SO4 --> BaSO4 +2HCl(1)

NaOH +H2SO4-->Na2SO4 +H2O(3)

theo (1) :nHCl=2nBaCl2=0,1(mol)

theo(2):nNaOH=nHCl=0,1(mol)

=> nNaOH(3)=2 -0,1=1,9(mol)

theo (3) :nH2SO4=nNaOH=1,9(mol)

=>mH2SO4=1,9.98=182,2(g)