Em tham khảo nha :

\(\begin{array}{l}
5)\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = \dfrac{m}{M} = \dfrac{{6,2}}{{31}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
\dfrac{{0,2}}{4} < \dfrac{{0,35}}{5} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = {n_{{O_2}}} - \dfrac{5}{4}{n_P} = 0,1mol\\
{m_{{O_2}d}} = n \times M = 0,1 \times 32 = 3,2g\\
b)\\
{n_{{P_2}{O_5}}} = \dfrac{{{n_P}}}{2} = 0,1mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,1 \times 142 = 14,2g\\
6)\\
a)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{13,5}}{{27}} = 0,5mol\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
\dfrac{{0,5}}{4} > \dfrac{{0,3}}{3} \Rightarrow Al\text{ dư}\\
{n_{A{l_d}}} = {n_{Al}} - \dfrac{4}{3}{n_{{O_2}}} = 0,1mol\\
{m_{A{l_d}}} = n \times M = 0,1 \times 27 = 2,7g\\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,2mol\\
{m_{A{l_2}{O_3}}} = n \times M = 0,2 \times 102 = 20,4g
\end{array}\)

 

$\text{Câu 5:}$

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

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

$\text{Phương trình:}$

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

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

$\to \text{ Kê theo số mol của P}$

$\text{a) Chất dư là} O_2$

$\to n_{O_2{dư}}=0,35-0,25=0,1\text{(mol)}$

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

$\text{b)}$

$m_{P_2O_5}=0,1.142=14,2\text{(g)}$

$\text{Câu 6:}$

$n_{Al}=\dfrac{13,5}{27}=0,5\text{(mol)}$

$n_{O_2}=\dfrac{6,72}{22,4}=0,3\text{(mol)}$

$\text{Phương trình:}$

$4Al + 3O_2 \xrightarrow{t^0} 2Al_2O_3$

$\text{Do:} \dfrac{0,5}{4}> \dfrac{0,3}{3}$

$\to \text{ Kê theo số mol của khí oxi}$

$\text{a) Vậy chất dư là Al}$

$n_{Al{dư}}=0,5-0,4=0,1\text{(mol)}$

$\to m_{Al{dư}}=0,1.102=10,2\text{(g)}$

$\text{b)}$

$n_{Al_2O_3}= 0,2.102=20,4\text{(g)}$