Em tham khaảo nha :

\(\begin{array}{l}
4)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{n_C} = \dfrac{m}{M} = \dfrac{{3,6}}{{12}} = 0,3mol\\
{n_{{O_2}}} = {n_C} = 0,3mol\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,2mol\\
{m_{KCl{O_3}}} = n \times M = 0,2 \times 122,5 = 24,5g\\
5)\\
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{m}{M} = \dfrac{{6,2}}{{31}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
\dfrac{{0,2}}{4} < \dfrac{{0,35}}{5} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = {n_{{O_2}}} - \dfrac{5}{4}{n_P} = 0,35 - \dfrac{5}{4} \times 0,2 = 0,1mol\\
{m_{{O_2}d}} = n \times M = 0,1 \times 32 = 3,2g\\
{V_{{O_2}d}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
b)\\
{P_2}{O_5}:\text{Diphotpho Pentaoxit}\\
{n_{{P_2}{O_5}}} = \dfrac{{{n_P}}}{2} = 0,1mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,1 \times 142 = 14,2g
\end{array}\)

 

$\text{BT4:}$

$n_C=\dfrac{3,6}{12}=0,3(mol)$

$\text{Phương trình:}$

$C + O_2 \xrightarrow{t^0} CO_2$

0,3 __     0,3   __ 0,3  $\text{(mol)}$

$2KClO_3 \xrightarrow{t^0} 2KCl + 3 O_2$

$\text{Ta có:} n_{O_2}=0,3(mol)$

$\to n_{KClO_2}=0,3.\dfrac{2}{3}=0,2(mol)$

$\to m_{KClO_3}=0,2.122,5=24,5(g)$

$\text{BT5:}$

$n_P=\dfrac{6,2}{31}=0,2(mol)$

$n_{O_2}=\dfrac{7,84}{22,4}=0,35(mol)$

$\text{Phương trình:}$

$4P +5O_2 \xrightarrow{t^0} 2P_2O_5$

$\text{Do:} \dfrac{0,2}{4}<\dfrac{0,35}{5}(mol)$

$\to \text{Kê theo số mol của P}$

$\text{a) Chất dư là:} O_2$

$\to n_{O_2\text{..dư}}=0,35-0,2.\dfrac{5}{4}=0,1(mol)$

$\to m_{O_2\text{..dư}}=0,1.32=3,2(g)$

$\text{b) Tên sản phẩm: điphotpho pentaoxit} (P_2O_5)$

$n_{P_2O_5}=\dfrac{0,2}{2}=0,1(mol)$

$\to m_{P_2O_5}=142.0,1=14,2(g)$