\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V=V_{O_2}=0,3\cdot22,4=6,72\left(l\right)\\a=m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)

a) $4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

b) $n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$n_{Al\ pư} = \dfrac{4}{3}n_{O_2} = 0,4(mol)$
$m_{Al\ pư} = 0,4.27 = 10,8(gam)$

c) 

Cách 1 : 

$m_{Al_2O_3} = m_{Al} + m_{O_2} = 10,8 + 0,3.32 = 20,4(gam)$

Cách 2 : 

Theo PTHH, $n_{Al_2O_3} = \dfrac{1}{2}n_{Al\ pư} = 0,2(mol)$
$m_{Al_2O_3} = 0,2.102 = 20,4(gam)$

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{O_2}=n.M=0,3.32=9,6\left(g\right)\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo định luật bảo toàn khối lượng, ta có:

\(m_{Al}+m_{O_2}=m_{Al_2O_3}\\ \rightarrow m_{Al_2O_3}=10,8+9,6=20,4\left(g\right)\)

a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

 0,6     0,45           0,3                ( mol )

\(m_{Al}=0,6.27=16,2g\)

\(V_{O_2}=0,45.22,4=10,08l\)

\(V_{kk}=10,08.5=50,4l\)

b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45 ( mol )

\(m_{KClO_3}=0,3.122,5=36,75g\)

c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45  ( mol )

\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)

\(m_{KClO_3}=0,4.122,5=49g\)

a/ PTHH

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(n_{O_2}=\dfrac{3,5}{22,4}=0,15625\left(mol\right)\)

Thep ĐLBTKL: m_M + m_O2 = m_MxOy

=> m_M = 8,875 - 0,15625.32 = 3,875(g)

=> \(n_M=\dfrac{3,875}{M_M}\left(mol\right)\)

PTHH: 2xM + yO₂ --t^o--> 2M_xO_y

___\(\dfrac{3,875}{M_M}\) ->\(\dfrac{3,875y}{2x.M_M}\)

=> \(\dfrac{3,875y}{2x.M_M}=0,15625=>M_M=\dfrac{62y}{5x}=\dfrac{31}{5}.\dfrac{2y}{x}\)

Xét \(\dfrac{2y}{x}=5\) => M_M = 31(P) => \(\dfrac{x}{y}=\dfrac{2}{5}\) => CTHH: P₂O₅

Áp dụng ĐLBTKL : 

m_Al + m_O2 = m_Al2O3 

8,1 + 4,032 : 22,4 × 32 = 13,86 (g)

\(n_{O_2} = \dfrac{4,032}{22,4} = 0,18(mol)\\ n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ \dfrac{n_{Al}}{4} = 0,075 < \dfrac{n_{O_2}}{3} = 0,06\)

Suy ra: Al dư

Bảo toàn khối lượng :

\(m = m_{Al\ dư} + m_{Al_2O_3} = m_{Al\ dư} + m_{Al\ pư} + m_{O_2}=m_{Al\ ban\ đầu} + m_{O_2} = 8,1 + 0,18.32 = 13,86(gam)\)

Cách mà mình hay dùng:

\(20,4-10,8=9,6\) 

\(\Rightarrow A\)

Cái này do mik nghĩ ra từ trên lớp nên ko cần áp dụng cthức gì đâu:)

\(PTHH:2Al_2O_3\underrightarrow{t^o}4Al+3O_2\)

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{Al_2O_3}=m_{Al}+m_{O_2}\\ m_{O_2}=m_{Al_2O_3}-m_{Al}=20,4-10,8=9,6\left(g\right)\)

\(n_{O_2}=\dfrac{9,6}{16}=0,6\left(mol\right)\\ V_{O_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)