Câu 7 :
Gọi $C_{M_{HCl}} = a(M) ; C_{M_{NaOH}} = b(M)$
Thí nghiệm 1 : $n_{HCl} = 0,3a(mol) ; n_{NaOH} = 0,1b(mol) ; n_{Ca(OH)_2} = 0,05.1 = 0,05(mol)$
$NaOH + HCl \to NaCl + H_2O$
$Ca(OH)_2 + 2HCl \to CaCl_2 + 2H_2O$
$n_{HCl} = n_{NaOH} + 2n_{Ca(OH)_2}$
$\Rightarrow 0,3a = 0,1b + 0,05.2(1)$
Thí nghiệm 2 : $n_{HCl} = 0,2a(mol) ; n_{NaOH} = 0,2b(mol); n_{H_2SO_4} = \dfrac{50.19,6\%}{98} = 0,1(mol)$
$NaOH + HCl \to NaCl + H_2O$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = n_{HCl} + 2n_{H_2SO_4}$
$\Rightarrow 0,2b = 0,2a + 0,1.2(2)$
Từ (1)(2) suy ra : a = 1(M) ; b = 2(M)
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