a,
Kim loại chu kì 4, nhóm IA là kali (K): x mol.
Kim loại chu kì 6, nhóm IIA là bari (Ba): y mol.
$\Rightarrow 39x+137y=5,86$ (1)
$n_{H_2}=\dfrac{1,344}{22,4}=0,06(mol)$
Bảo toàn e: $x+2y=0,06.2=0,12$ (2)
$(1)(2)\Rightarrow x=0,08; y=0,02$
$m_K=39x=3,12g$
$m_{Ba}=137y=2,74g$
b,
$m_X=5,86+44,26-0,06.2=50g$
$n_{KOH}=x=0,08(mol)$
$\to C\%_{KOH}=\dfrac{0,08.56.100}{50}=8,96\%$
$n_{Ba(OH)_2}=y=0,02(mol)$
$\to C\%_{Ba(OH)_2}=\dfrac{0,02.171.100}{50}=6,84\%$
$V_X=50:1,25=40ml=0,04l$
$\Rightarrow C_{M_{KOH}}=\dfrac{0,08}{0,04}=2M; C_{M_{Ba(OH)_2}}=\dfrac{0,02}{0,04}=0,5M$
$2KOH+H_2SO_4\to K_2SO_4+2H_2O$
$Ba(OH)_2+H_2SO_4\to BaSO_4+2H_2O$
$\Rightarrow n_{H_2SO_4}=0,5x+y=0,06(mol)$
$\to V_{H_2SO_4}=\dfrac{0,06}{0,3}=0,2l$