$CH_4+2O_2 \rightarrow CO_2 +2H_2O$$*TH_1:$
$a/$ $CO_2+Ba(OH)_2\rightarrow BaCO_3+H_2O$
$0,07$ $0,07$
$\Rightarrow n_{CH_4}=0,07 \Rightarrow V_{CH_4}=0,07\times 22,4=1,568 (l)$
$b/$ $n_{H_2O}=0,07\times 2=0,14(mol)$
$m_{tăng}=0,7\times 44+0,14\times 18=7,6(g)$
$c/$ $\Delta m=0,07\times44+0,14\times18-17,75=-10,15(g)$
$\Rightarrow$ giảm $10,15 g$
$*TH_2:$
$a/$ $CO_2+Ba(OH)_2\rightarrow BaCO_3+H_2O$
$0,07$ $0,07$
$2CO_2+Ba(OH)_2\rightarrow Ba(HCO_3)_2$
$0,06$ $0,03$
$\Rightarrow n_{CH_4}=0,07+0,06=0,13(mol) \Rightarrow V=0,13\times22,4=2,912$
$b/$ $m_{tăng}=0,13\times44+0,13\times2\times18=10,4$
$c/$ $\Delta m=10,4-15,75=-5,35$
$\Rightarrow$ Giảm $5,35 (g)$