Đáp án:

`↓`

Giải thích các bước giải:

`n_C``=``\frac{m}{M}``=``\frac{3}{12}``=``0,25` `(mol)`

`PTHH`               `C``+``O_2` $\xrightarrow[]{t^o}$ `CO_2↑`

                          `0,25`  `0,25`                     `mol`

`→``n_{O_2}``=``n_C``=``0,25` `(mol)`

`⇒``m_{O_2}``=``n``.``M``=``0,25``.``32``=``8` `(g)`

`⇒``V_{O_2}``=``n``.``22,4``=``0,25``.``22,4``=``5,6` `(l)`

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$n_{C_4H_10(đktc)}$`=``\frac{V}{22,4}``=``\frac{11,2}{22,4}``=``0,5` `(mol)`

`PTHH`               `2C_4H_10``+``13O_2` $\xrightarrow[]{t^o}$ `8CO_2↑``+``10H_2O`

                                  `0,5`            `3,25`                                      `mol`

`→``n_{O_2}``=``n_{C_4H_10}``=``\frac{0,5.13}{2}``=``3,25` `(mol)`

`⇒``m_{O_2}``=``n``.``M``=``3,25``.``32``=``104` `(g)`

`⇒``V_{O_2}``=``n``.``22,4``=``3,25``.``22,4``=``72,8` `(l)`

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`n_P``=``\frac{m}{M}``=``\frac{0,62}{31}``=``0,02` `(mol)`

`PTHH`           `4P``+``5O_2` $\xrightarrow[]{t^o}$ `2P_2O_5`

                      `0,02`    `0,025`                      `mol`

`→``n_{O_2}``=``n_P``=``\frac{0,02.5}{4}``=``0,025` `(mol)`

`⇒``m_{O_2}``=``n``.``M``=``0,025``.``32``=``0,8` `(g)`

`⇒``V_{O_2}``=``n``.``22,4``=``0,025``.``22,4``=``0,56` `(l)`

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`n_{CO}``=``\frac{m}{M}``=``\frac{14}{28}``=``0,5` `(mol)`

`PTHH`            `2CO``+``O_2` $\xrightarrow[]{t^o}$ `2CO_2`

                           `0,5`     `0,25`                       `mol`

`→``n_{O_2}``=``n_{CO}``=``\frac{0,5.1}{2}``=``0,25` `(mol)`

`⇒``m_{O_2}``=``n``.``M``=``0,25``.``32``=``8` `(g)`

`⇒``V_{O_2}``=``n``.``22,4``=``0,25``.``22,4``=``5,6` `(l)`

Đáp án + Giải thích các bước giải:

Câu 1:

$n_C=\dfrac{3}{12}=0,25(mol)$

$n_{C_4H_{10}}=\dfrac{11,2}{22,4}=0,5(mol)$

$n_P=\dfrac{0,62}{31}=0,02(mol)$

$n_{CO}=\dfrac{14}{28}=0,5(mol)$

Các PTHH của từng trường hợp:

$C+O_2\xrightarrow{t^o}CO_2(1)$

$2C_4H_{10}+13O_2\xrightarrow{t^o}8CO_2\uparrow+10H_2O(2)$

$4P+5O_2\xrightarrow{t^o}2P_2O_5(3)$

$2CO+O_2\xrightarrow{t^o}2CO_2(4)$

Với PT(1): $n_{O_2}=n_C=0,25(mol)$

$\Rightarrow \begin{cases} m_{O_2}=0,25.32=8(g)\\ V_{O_2}=0,25.22,4=5,6(lít) \end{cases}$

Với PT(2): $n_{O_2}=6,5n_{C_4H_{10}}=3,25(mol)$

$\Rightarrow \begin{cases} m_{O_2}=3,25.32=104(g)\\ V_{O_2}=3,25.22,4=72,8(lít) \end{cases}$

Với PT(3): $n_{O_2}=1,25n_P=0,025(mol)$

$\Rightarrow \begin{cases} m_{O_2}=0,025.32=0,8(g)\\ V_{O_2}=0,025.22,4=0,56(lít) \end{cases}$

Với PT(4): $n_{O_2}=0,5n_{CO}=0,25(mol)$

$\Rightarrow \begin{cases} m_{O_2}=0,25.32=8(g)\\ V_{O_2}=0,25.22,4=5,6(lít) \end{cases}$