\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ a)4Al+3O_2\xrightarrow[t^0]{}2Al_2O_3\\ 0,2........0,15......0,1\\ b)m_{Al_2O_3}=0,1.102=10,2g\\ c)V_{O_2}=0,15.24,79=3,7185l\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

                       4          3             2

                      0,2      0,15         0,1 (mol)

\(b)m_{Al_2O_3}=n\cdot M=0,1\cdot\left(27\cdot2+16\cdot3\right)=10,2\left(g\right)\\ c)V_{O_2}=n\cdot24,79=0,15\cdot24,79=3,7185\left(l\right).\)

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

Ta có : 

\(n_{Al_2O_3}=\dfrac{0.2\cdot2}{4}=0.1\left(mol\right)\)

\(m_{Al_2O_3}=0.1\cdot102=10.2\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.2.......\dfrac{2}{15}.....\dfrac{1}{15}\)

\(V_{O_2}=\dfrac{2}{15}\cdot22.4=2.98\left(l\right)\)

\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.46\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,2}{4}< \dfrac{0,2}{3}\Rightarrow O_2dư\\ \Rightarrow n_{O_2\left(dư\right)}=0,2-\dfrac{3}{4}.0,2=0,05\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\\ b,n_{Al_2O_3}=\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,1=10,2\left(g\right)\)

a, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,2}{3}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Zn}=0,2mol\\ a.2Zn+O_2-^{^{ }t^{^0}}->2ZnO\\ b.m_{ZnO}=0,2.71=14,2g\\ n_{O_2}=0,2:2=0,1mol\\ V_{O_2}=0,1.22,4=2,24L\\ c.2KClO_3-^{^{ }t^{^{ }0}}->2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,1=\dfrac{0,2}{3}mol\\ m_{KClO_3}=122,5\cdot\dfrac{0,2}{3}=8,166g\)

Giúp e lẹ với ạ

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2                            0,1                0,3  ( mol )

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(V_{H_2}=0,3.22,4=6,72l\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:2Al+6HCl->2AlCl_3+3H_2\)

tỉ lệ          2    :       6       :        2     :     3

n(mol)     0,2---->0,6------->0,2------------->0,3

\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\\ m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

a) 

4Al + 3O₂ --t^o--> 2Al₂O₃

b) \(n_{O_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

            0,2<-0,15------->0,1

=> m_Al = 0,2.27 = 5,4 (g)

c) m_Al2O3 = 0,1.102 = 10,2 (g)

a) 4Al + 3O₂ --t^o--> 2Al₂O₃

b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

           0,4-->0,3-------->0,2

V_O2(đkc) = 0,3.24,79 = 7,437 (l)

c) m_Al2O3 = 0,2.102 = 20,4 (g)

\(PTHH:4Al+3O_2->2Al_2O_3\)

BĐ          0,4      0,27                     (mol)

PU        0,36---->0,27---->0,18         (mol) 

CL          0,04---->0------>0,18     (mol)

b)

\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\)

\(\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\left(\dfrac{0,4}{4}>\dfrac{0,27}{3}\right)\)

=> Al dư, O2 hết (tính theo O2)

\(m_{Al}=n\cdot M=0,04\cdot27=1,08\left(g\right)\)

c)

\(m_{Al_2O_3}=n\cdot M=0,18\cdot\left(27\cdot2+16\cdot3\right)=18,36\left(g\right)\)

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

b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{4}>\dfrac{0,27}{3}\), ta được Al dư.

Theo PT: \(n_{Al\left(pư\right)}=\dfrac{4}{3}n_{O_2}=0,36\left(mol\right)\)

\(\Rightarrow n_{Al\left(dư\right)}=0,4-0,36=0,04\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,04.27=1,08\left(g\right)\)

c, Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{Al}=0,18\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,18.102=18,36\left(g\right)\)