Câu 1:
$n_{H_2SO_4}=0,2.1=0,2(mol)$
a,
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\to n_{NaOH}=2n_{H_2SO_4}=0,4(mol)$
$\to m_{dd NaOH}=0,4.40:20\%=80g$
b,
$2KOH+H_2SO_4\to K_2SO_4+2H_2O$
$\to n_{KOH}=2n_{H_2SO_4}=0,4(mol)$
$\to V_{dd KOH}=0,4.56:5,6\%:1,045=382,775ml$
Câu 2:
$n_{HCl}=0,2.0,2=0,04(mol)$
a,
$NaOH+HCl\to NaCl+H_2O$
$\to n_{NaOH}=n_{NaCl}=n_{HCl}=0,04(mol)$
$\to V_{dd NaOH}=\dfrac{0,04}{0,1}=0,4l=400ml$
$V_{dd\text{spu}}=0,4+0,2=0,6l$
$\to C_{M_{NaCl}}=\dfrac{0,04}{0,6}=\dfrac{1}{15}(M)$
b,
$2HCl+Ca(OH)_2\to CaCl_2+2H_2O$
$\to n_{Ca(OH)_2}=n_{CaCl_2}=\dfrac{n_{HCl}}{2}=0,02(mol)$
$\to m_{dd Ca(OH)_2}=0,02.74:5\%=29,6g$
$m_{dd HCl}=200.1=200g$
$\to m_{dd\text{spu}}=29,6+200=229,6g$
$\to C\%_{CaCl_2}=\dfrac{0,02.111.100}{229,6}=0,97\%$
