3Fe+2O2-to>Fe3O4

0,1----1\15--------1\30

n Fe=\(\dfrac{5.6}{56}\)=0,1 mol

n O2=\(\dfrac{2,24}{22,4}\)=0,1 mol

=>O2 dư

=>m O2du=(0,1-\(\dfrac{1}{15}\)).32=1,067g

=>m Fe3O4=\(\dfrac{1}{30}.232=7,73g\)

\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)

\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)

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

\(3.........2\)

\(0.225......0.1875\)

Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)

\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)

\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)

\(4Na+O_2\underrightarrow{^{^{t^0}}}2Na_2O\)

\(4..........1\)

\(0.2.....0.02\)

\(LTL:\dfrac{0.2}{4}>\dfrac{0.02}{1}\Rightarrow Nadư\)

\(m_{Na\left(dư\right)}=\left(0.2-0.08\right)\cdot23=2.76\left(g\right)\)

\(m_{Na_2O}=0.04\cdot62=2.48\left(g\right)\)

nP= 7,44/31=0,24(mol)

nO2=6,16/22,4=0,275(mol)

PTHH:4 P + 5 O2 -to->2  P2O5

Ta có: 0,24/4 > 0,275/5

=> O2 hết, P dư, tính theo nO2

nP(p.ứ)= 0,275 x 4/5= 0,22(mol)

=>nP(dư)=0,24-0,22=0,02(mol)

=>mP(dư)=0,02.31= 0,62(g)

nP2O5= 2/5 x 0,275= 0,11(mol)

=> mP2O5= 142 x 0,11= 15,62(g)

\(n_P=\dfrac{7,44}{31}=0,24\left(mol\right)\)

\(n_{O_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)

PTHH :                       \(4P+5O_2\rightarrow2P_2O_5\)

Ban đầu :                0,24    0,275                   (mol)

Phản ứng :             0,22    0,275       0,11    (mol)

Sau phản ứng :      0,02      0            0,11    (mol)

\(m_P=0,02.31=0,62\left(g\right)\)

\(m_{P_2O_5}=0,11.142=15,62\left(g\right)\) 

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{12,8}{32}=0,4\left(mol\right)\)

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

tỉ lệ:            4  :  3     :       2

n(mol)       0,2   0,4

m(mol p/u) 0,2-->0,15---->0,1

\(\dfrac{n_{Al}}{4}< \dfrac{n_{O_2}}{3}\left(\dfrac{0,2}{4}< \dfrac{0,4}{3}\right)\)

`=>` `Al` hết , `O_2` dư

`=>` tính theo `Al`

\(n_{O_2\left(dư\right)}=0,4-0,15=0,25\left(mol\right)\\ m_{O_2\left(dư\right)}=n\cdot M=0,25\cdot32=8\left(g\right)\\ m_{Al_2O_3}=n\cdot M=0,1\cdot\left(27\cdot2+16\cdot3\right)=10,2\left(g\right)\)

Giúp ik

a)

\(n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_P = \dfrac{4}{5}n_{O_2} = 0,4(mol)\\ \Rightarrow m_P = 0,4.31 = 12,4(gam)\)

b)

\(n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\)

c)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)

bạn từng câu lên sẽ dễ nhìn hơn