\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3.0,2}{4}=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow m_{KMnO_4}=158.0,3=47,4\left(g\right)\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{19,5}{65}=0,3mol\)

\(Zn+\dfrac{1}{2}O_2\rightarrow\left(t^o\right)ZnO\)

1      1/2                   1       (mol)

0,3     0,15                   0,3        ( mol )

PƯ trên thuộc loại phản ứng hóa hợp

\(m_{ZnO}=n_{ZnO}.M_{ZnO}=0,3.81=24,3g\)

\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)

a, 3Fe + 2O₂ -t^o-> Fe₃O₄

b, n_Fe = m/M = 16,8/56 = 0,3 (mol)

từ pthh ta có: \(n_{O_2}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)

=>\(V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

c, C1: từ pthh ta có: \(n_{Fe_3O_4}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)

=>\(m_{Fe_3O_4}=n.M=0,1.\left(56.3+4.16\right)=0,1.232=23,2\left(g\right)\)

C2: \(m_{O_2}=n.M=0,2.32=6,4\left(g\right)\)

Áp dụng ĐLBTKL ta co:

\(m_{Fe_3O_4}=m_{Fe}+m_{O_2}=16,8+6,4=23,2\left(g\right)\)

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

a. \(PTHH:4P+5O_2\overset{t^o}{--->}2P_2O_5\)

Theo PT: \(n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,25.22,4=5,5\left(lít\right)\)

b. Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ 4P+5O_2-^{t^o}\rightarrow2P_2O_5\\ a.n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\\ \Rightarrow m_{O_2}=0,25.32=8\left(g\right)\\ b.BTKLm_P+m_{O_2}=m_{P_2O_5}\\ \Rightarrow m_{P_2O_5}=6,2+8=14,2\left(g\right)\)

Bài 4 câu a đề là thể tích H2 nha bạn

a)\(Fe2O3+3H2-->2Fe+3H2O\)

\(n_{Fe2O3}=\frac{12}{160}=0,075\left(mol\right)\)

\(n_{H2}=3n_{Fe2O3}=0,225\left(mol\right)\)

\(V_{H2}=0,225.22,4=5,04\left(l\right)\)

b)\(n_{Fe}=2n_{Fe2O3}=0,15\left(mol\right)\)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

Bài 6

a)\(Zn+H2SO4-->ZnSO4+H2\)

\(n_{Zn}=\frac{19,5}{65}=0,3\left(mol\right)\)

\(n_{ZnSO4}=n_{Zn}=0,3\left(mol\right)\)

\(m_{ZnSO4}=0,3.162=48,3\left(g\right)\)

b)\(n_{H2}=n_{Zn}=0,3\left(mol\right)\)

\(V_{H2}=0,3.22,4=6,72\left(l\right)\)

a)Fe2O3+3H2−−>2Fe+3H2O

0,075-------0,225---0,15 mol

nFe2O3=12/160=0,075(mol)

VH2=0,225.22,4=5,04(l)

b)

mFe=0,15.56=8,4(g)

Bài 6

a)Zn+H2SO4−−>ZnSO4+H2

0,3----------------------0,3---0,3

nZn=19,5/65=0,3(mol)

mZnSO4=0,3.162=48,3(g)

b)

=>VH2=0,3.22,4=6,72(l)

:V

hình như viết chữ hơi cẩu thả một tí:>

chữ đẹp ='')

a/ Ta có: \(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

PTHH: 

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

2          1

0.2       x

\(=>x=\dfrac{0.2\cdot1}{2}=0.1=n_{O_2}\)  

\(=>V_{O_2\left(đktc\right)}=0.1\cdot22.4=2.24\left(l\right)\)

b/ \(2Mg+O_2\underrightarrow{t^o}2MgO\)

     2                  2

     0.2               y

\(=>y=\left(0.2\cdot2\right):2=0.2=n_{MgO}\)

\(=>m_{MgO}=0.2\cdot\left(24+16\right)=8\left(g\right)\)

mình cảm ơn bạn ạ

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

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

0,2      0,15    0,1

\(m_{Al_2O_3}=0,1\cdot102=10,2g\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

0,1                            0,15

\(m_{KClO_3}=0,1\cdot122,5=12,25g\)

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

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

\(\Rightarrow n_{Al_2O_3}=0,2.2:4=0,1\left(mol\right);n_{O_2}=0,2.3:4=0,15\left(mol\right)\)

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

b)\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(n_{O_2}=0,15\left(mol\right)\)(câu a)

\(\Rightarrow n_{KClO_3}=0,15.2:3=0,1\left(mol\right)\)

\(m_{KClO_3}=0,1.123,5=12,35\left(g\right)\)

nFe= 2,8/56=0,05(mol)

PTHH: 3Fe+2O2-> Fe3O4

           0,05   0,003   0,0167

a) mFe3O4= 0,0167. 232=3,8744(g)

b) Vo2 = 0,003.22,4=0,0672(l)

Ta có : nFe=2,8/56=0,05 mol

3Fe+2O2--->Fe3O4 (1)

a) theo pt (1) nFe3O4=1/3nFe=1/60 mol 

=> mFe3O4=1/60 . MFe3O4 =58/15 gam

b) nO2 =2/3 nFe=0,05 .2/3=1/30 mol

=> VO2=1/30 .22,4=0,7467 lít 

c) Vkhông khí=5. VO2=3,73 lít.