Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
b)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_P} = \dfrac{4}{5}{n_{{O_2}}} = 0,24mol\\
{m_P} = n \times M = 0,24 \times 31 = 7,44g\\
c)\\
{n_{{P_2}{O_5}}} = \dfrac{2}{5}{n_{{O_2}}} = 0,12mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,12 \times 142 = 17,04mol\\
2)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{21,6}}{{27}} = 0,8mol\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,4mol\\
{m_{A{l_2}{O_3}}} = n \times M = 0,4 \times 102 = 40,8g\\
c)\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Al}} = 0,6mol\\
{V_{{O_2}}} = n \times 22,4 = 0,6 \times 22,4 = 13,44l
\end{array}\)

 

bài 1:

a) 4P + 5$O_{2}$ → 2$P_{2}$ $O_{5}$ 

b) $n_{O_{2} }$=$\frac{V}{22,4}$ =$\frac{6,72}{22,4}$ =0,3 mol

4P + 5$O_{2}$ → 2$P_{2}$ $O_{5}$ 

  4         5                       2        (mol)

0,24       0,3                   0,12    (mol)

$m_{P}$=n.M=0,24.31=7,44g

c) $m_{P_{2}O_{5} }$ =n.M=0,12.142=17,04g

bài 2:

a) 4Al + 3$O_{2}$ → 2$Al_{2}$ $O_{3}$ 

b) $n_{Al}$ = $\frac{m}{M}$ =$\frac{21,6}{27}$ =0,8 mol

4Al + 3$O_{2}$ → 2$Al_{2}$ $O_{3}$ 

4            3                     2          (mol)

0,8          0,6                  0,4        (mol)

$m_{Al_{2}O_{3} }$=n.M=0,4.102=40,8g

c) $V_{O_{2}}$ =n.22,4=0,6.22,4=13,44l