a. \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

PTHH : 2Mg + O₂ ----t^o----> 2MgO

              0,5       0,25             0,5

b. \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

c. \(m_{MgO}=0,5.40=20\left(g\right)\)

nMg = 12 : 24 = 0,5 (mol) 
pthh : 2Mg + O2-t-> 2MgO
          0,5     0,25           0,5
mMgO = 0,5. 40 = 20 (G)
V_O2 = 0,25. 22,4 = 5,6 (G)

\(n_{Mg}=\dfrac{12}{24}=0,5mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

0,5       0,25            0,5        ( mol )

\(V_{O_2}=0,25.22,4=5,6l\)

\(m_{MgO}=0,5.40=20g\)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

            0,4<--0,5------->0,2

=> m_P2O5 = 0,2.142 = 28,4 (g)

c_ m_P = 0,4.31 = 12,4 (g)

\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

PTHH: 4P + 5O₂ --to--> 2P₂O₅

           0,4      0,5             0,2

\(\rightarrow\left\{{}\begin{matrix}m_{P_2O_5}=0,2.142=28,4\left(g\right)\\m_P=0,4.31=12,4\left(g\right)\end{matrix}\right.\)

\(n_C=\dfrac{0,6}{12}=0,05mol\)

\(n_{O_2}=\dfrac{1,6}{32}=0,05mol\)

\(C+O_2\rightarrow\left(t^o\right)CO_2\)

0,05 = 0,05                  ( mol )

0,05   0,05        0,05    ( mol )

\(m_{CO_2}=0,05.44=2,2g\)

\(\dfrac{M_{CO_2}}{M_{H_2}}=\dfrac{44}{2}=22\)

=> Khí thu được nặng hơn khí hiđro 22 lần

\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

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

\(\left\{{}\begin{matrix}m_{O_2}=0,2.32=6,4g\\V_{O_2}=0,2.22,4=4,48l\end{matrix}\right.\)

\(m_{Fe_3O_4}=0,1.232=23,2g\)

REFER

a)3 Fe+2O2--->Fe3O4

b) Ta có

n Fe=16,8/56=0,3(mol)

Theo pthh

n O2=2/3n Fe=0,2(mol)

V O2=0,2.22,4=4,48(l)

c) Cách 1

Áp dụng định luật bảo toàn khối lượng ta có

m Fe3O4=m Fe+m O2

=16,8+0,2.32=23,2(g)

Cách 2

Theo pthh

n Fe3O4=1/3n Fe=0,1(mol)

m Fe3O4=0,1.232=23,2(g)

\(n_P=\dfrac{12,4}{31}=0,4mol\)

\(n_{O_2}=\dfrac{17}{32}=0,53125mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

\(\dfrac{0,4}{4}\) < \(\dfrac{0,53125}{5}\)                ( mol )

0,4        0,5             0,2       ( mol )

\(m_{O_2\left(dư\right)}=\left(0,53125-0,5\right).32=1g\)

\(m_{P_2O_5}=0,2.142=28,4g\)

\(a,n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O4

LTL: \(\dfrac{0,2}{3}>\dfrac{0,1}{2}\) => Fe dư

b, Theo pthh: \(n_{Fe\left(pư\right)}=\dfrac{3}{2}n_{O_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=\left(0,2-0,15\right).56=2,8\left(g\right)\)

4Al+3O2-to>2Al2O3

0,2-----0,15-----0,1 mol

n Al=0,2 mol

=>m O2=0,15.32=4,8g

=>Vkk=0,15.22,4.5=16,8l

=>m Al2O3=0,1.102=10,2g

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

\(BTKL:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\)

\(\Leftrightarrow m_{Fe_3O_4}=8.4+\dfrac{4.48}{22.4}\cdot32=14.8\left(g\right)\)

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

b)\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

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

Lập tỉ lệ ta có : \(\dfrac{n_{Fe}}{3}< \dfrac{n_{O_2}}{2}\)

=> Sau phản ứng O2 dư, Fe phản ứng hết

Chất rắn sau phản ứng là Fe3O4

\(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,05\left(mol\right)\)

=> m_Fe3O4=0,05.232=11,6g

  PTHH: 3Fe + 2O₂ ------> Fe₃O₄ 

nO2 = 4.48 : 22.4 = 0.2  (mol)

mO2 = 0.2 . 32 = 6,4 (g)

Áp dụng Định luật BTKL ta được: mFe + mO2 = mFe₃O₄  

                                                         8,4 + 6,4   =  mFe₃O₄  

=>  mFe₃O₄   = 14,8 (g)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{50,4}{2.22,4}=0,45\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

LTL: \(\dfrac{0,2}{4}< \dfrac{0,45}{5}\rightarrow\) O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\n_{O_2\left(pư\right)}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,45-0,15\right).32=9,6\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)