Đáp án:
`a) 1,12(l)`
`b) 5,6(l)`
Giải thích các bước giải:
`n_{S}``=``\frac{m}{M}``=``\frac{1,6}{32}``=0,05 (mol)`
`PTHH`: `S``+O_{2} `$\underrightarrow{t^0}$$SO_{2}$
`⇒` `n_{O_2}=0,05` (mol)
`⇒` `n_{SO_2}=0,05` (mol)
`a)`$V_S=n.22,4=0,05.22,4=1,12\left(l\right)$
`b) PTHH:` $O_2+2H_2\underrightarrow{t^0}2H_2O$
`⇒` `n_{O_2}=0,5` (mol)
`⇒` `n_{H_2}=0,1` (mol)
`⇒` `n_{H_2O}=0,1` (mol)
`⇒` $V_{0_2}=n.22,4=0,05.22,4=1,12\left(l\right)$
`⇒` $V_{KK}=5.V_{O_2}=5.1,12=5,6\left(l\right)$