`pt:` `2Cu`+`O2`--t*->`2CuO`

`2Mg`+`O2`--->`2MgO`

b) `nO2`=`{3,36}/22,4` =`0,15`(`mol`)

`2KClO3`$\Rightarrow$`2KCl+3O2`

`0,1`______________`0,15`

`=>``mKClO3`=`0,1.122,5`=`12,25`(`g`)

$\textit{Noicomdien1}$

 

$\text{Đáp án+Giải thích các bước giải:}$

$\text{A.}$

$\text{2Cu + $O_{2}$$\xrightarrow{t^o}$2CuO (1)}$

$\text{2Mg + $O_{2}$$\xrightarrow{t^o}$2MgO}$

$\text{B.}$

$\text{Gọi x, y lần lượt là số mol của Cu, Mg}$

$\text{Ta có: $m_{HH}$=$m_{Cu}$ + $m_{Mg}$}$

$\text{Mà: $m_{Cu}$= 64x}$

$\text{      $m_{Mg}$= 24y}$

$\text{⇒ 64x + 24y = 11,2 (*)}$

$\text{Ta có: $n_{O_{2}}$=$\dfrac{3,36}{22,4}$= 0,15 (mol)}$

$\text{Theo PT (1): $n_{O_{2}}$=$\dfrac{1}{2}$$n_{Cu}$= $\dfrac{1}{2}$x (mol)}$

$\text{Theo PT (2): $n_{O_{2}}$=$\dfrac{1}{2}$$n_{Mg}$= $\dfrac{1}{2}$y (mol)}$

$\text{⇒ $\dfrac{1}{2}$x + $\dfrac{1}{2}$y= 0,15 (**)}$

$\text{Từ (*), (**) ta có HPT:}$

$\begin{cases} 64x + 24y = 11,2\\\dfrac{1}{2}x + \dfrac{1}{2}y= 0,15 \end{cases}$

⇒ $\begin{cases} x= 0,1\\y= 0,2 \end{cases}$

$\text{⇒ $m_{Cu}$=0,1 × 64= 6,4 (g)}$

$\text{    $m_{Mg}$=0,2 × 24= 4,8 (g)}$

$\text{C.}$

$\text{$2KClO_{3}$$\xrightarrow{t^o}$2KCl + $3O_{2}$ (3)}$

$\text{Theo PT (3): $n_{KClO_{3}}$=$\dfrac{2}{3}$$n_{O_{2}}$=$\dfrac{2}{3}$ × 0,15= 0,1 (mol)}$

$\text{⇒ $m_{KClO_{3}}$=0,1 × 122,5= 12,25 (g)}$