$n_{Ba(OH)_2}=0,5.0,2=0,1(mol)$
$n_{BaCO_3}=\dfrac{15,76}{197}=0,08(mol)<0,1(mol)$
* Trường hợp 1: dư $Ba(OH)_2$
$CH_4+2O_2\buildrel{{t^o}}\over\to CO_2+H_2O$
$Ba(OH)_2+CO_2\to BaCO_3+H_2O$
a,
$n_{CH_4}=n_{CO_2}=n_{BaCO_3}=0,08(mol)$
$\to V=0,08.22,4=1,792l$
b,
$n_{H_2O}=2n_{CO_2}=0,16(mol)$
$\to m_{\text{bình tăng}}=m_{CO_2}+m_{H_2O}=0,08.44+0,16.18=6,4g$
c,
$\Delta m_{dd}=m_{CO_2}+m_{H_2O}-m_{BaCO_3}=0,08.44+0,16.18-15,76=-9,36g$
Vậy khối lượng dd giảm $9,36g$
* Trường hợp 2: 2 muối
a,
$CH_4+2O_2\buildrel{{t^o}}\over\to CO_2+2H_2O$
$2CO_2+Ba(OH)_2\to Ba(HCO_3)_2$
$CO_2+Ba(OH)_2\to BaCO_3+H_2O$
$n_{Ba(HCO_3)_2}=0,1-0,08=0,02(mol)$
$\to n_{CH_4}=n_{CO_2}=0,02.2+0,08=0,12(mol)$
$\to V=2,688l$
b,
$n_{H_2O}=2n_{CO_2}=0,24(mol)$
$\to m_{\text{bình tăng}}=0,12.44+0,24.18=9,6g$
c,
$\Delta m_{dd}=0,12.44+0,24.18-15,76=-6,16g$
Vậy khối lượng dd giảm $6,16g$
