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

b)

\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)

Theo PTHH : 

\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)

c)

\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)

PTHH: \(Mg+\dfrac{1}{2}O_2\xrightarrow[]{t^o}MgO\)

            \(Cu+\dfrac{1}{2}O_2\xrightarrow[]{t^o}CuO\)

Theo đề bài: \(m_{tăng}=32\left(g\right)=m_{O_2\left(p.ứ\right)}\)

Bảo toàn khối lượng: \(m_{oxit\:}=m_{KL}+m_{O_2}=120\left(g\right)\) 

\(\Rightarrow m_{MgO}=\dfrac{120}{3}=40\left(g\right)\) \(\Rightarrow m_{CuO}=80\left(g\right)\)

\(m_C=10.36\%=3,6\left(g\right)\Rightarrow n_C=0,3\left(mol\right)\)

\(m_S=10-3,6=6,4\left(g\right)\Rightarrow n_S=0,2\left(mol\right)\)

\(C+O_2-^{t^o}\rightarrow CO_2\)

\(S+O_2-^{t^o}\rightarrow SO_2\)

\(n_{O_2}=0,3+0,2=0,5\left(mol\right)\)

Vì oxi chiếm 20% thể tích không khí 

=> \(V_{kk}=\dfrac{0,5.22,4}{20\%}=56\left(lít\right)\)

\(n_{hh}=n_{CO_2}+n_{SO_2}=0,3+0,2=0,5\left(mol\right)\)

=> \(V_{hh}=0,5.22,4=11,2\left(lít\right)\)

PTHH: 

mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)

VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)

mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)

\(a,m_C=10.36\%=3,6\left(kg\right)=3600\left(g\right)\\ n_C=\dfrac{3600}{12}=300\left(mol\right)\\ m_S=10-3,6=6,4\left(kg\right)=6400\left(g\right)\\ n_S=\dfrac{6400}{32}=200\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ n_{O_2\left(tổng\right)}=n_C+n_S=300+200=500\left(mol\right)\\ V_{O_2\left(tổng\right)\left(đktc\right)}=500.22,4=11200\left(l\right)\\ V_{kk}=\dfrac{100}{20}V_{O_2\left(tổng\right)\left(đktc\right)}=5.11200=56000\left(l\right)\\ b,V_{hh\left(CO_2,SO_2\left(đktc\right)\right)}=22,4.\left(n_C+n_S\right)=22,4.\left(300+200\right)=11200\left(l\right)\)

a) 2Mg + O₂ --t^o--> 2MgO

4Al + 3O₂ --t^o--> 2Al₂O₃

b) Gọi số mol Mg, Al là a, b

=> 24a + 27b = 7,8 

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

PTHH: 2Mg + O₂ --t^o--> 2MgO

______a--->0,5a-------->a

4Al + 3O₂ --t^o--> 2Al₂O₃

b-->0,75b------->0,5b

=> 0,5a + 0,75b = 0,2

=> a = 0,1 ; b = 0,2

=> m_Mg = 0,1.24 = 2,4 (g); m_Al = 0,2.27 = 5,4 (g)

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{2,4}{7,8}.100\%=30,769\%\\\%Al=\dfrac{5,4}{7,8}.100\%=69,231\%\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}n_{MgO}=0,1\left(mol\right)\\n_{Al_2O_3}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)

=> m = 4 + 10,2 = 14,2 (g)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5mol\)

PTHH:

\(2Mg+O_2->2MgO\)

2     :      1      :      2          mol

1     :     0,5      :    1          mol

\(m_{Mg}=n.M=1.24=24g\)

\(m_{MgO}=n.M=1.\left(24+16\right)=40g\)

nO2 = 11,2/22,4 = 0,5 (mol)

PTHH: 2Mg + O2 -> (t°) 2MgO

Mol: 1 <--- 0,5 ---> 1

mMg = 1 . 24 = 24 (g)

mMgO = 1 . 40 = 40 (g)

a) \(n_{O_2}=\dfrac{11,2.20\%}{22,4}=0,1\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

            0,2<--0,1--------->0,2

=> m_Mg = 0,2.24 = 4,8 (g)

b) n_MgO = 0,2.40 = 8 (g)

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(2Mg+O_2\underrightarrow{t^0}2MgO\)

\(0.2.......0.1........0.2\)

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

\(m_{MgO}=0.2\cdot40=8\left(g\right)\)

a) 

2Mg + O2 --t^o--> 2MgO

0,2----->0,1------>0,2     (mol)

n_Mg = 4,48/24 = 0,2 (mol)

=> V_O2 = 0,1.22,4 = 2,24 (lít)

b) m_MgO = 0,2.(24 + 16) = 8 (g) 

\(m_{Al}=27,8.19,2\%=5,4\left(g\right)\\ m_{Fe}=27,8-5,4=22,4\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\end{matrix}\right.\)

PTHH:

4Al + 3O₂ --t^o--> 2Al₂O₃

0,2-->0,15------->0,1

3Fe + 2O₂ --t^o--> Fe₃O₄

0,4-->4/15--------->2/15

\(\rightarrow\left\{{}\begin{matrix}V_{kk}=\left(0,15+\dfrac{4}{15}\right).22,4.5=\dfrac{140}{3}\left(l\right)\\m_{Cran}=0,1.102+\dfrac{2}{15}.232=\dfrac{617}{15}\left(g\right)\end{matrix}\right.\)

mAl=27,8.19,42%=5,4g

⇒nAl=\(\dfrac{5,4}{27}\)=0,2mol

⇒nFe=\(\dfrac{27,8-5,4}{56}\)=0,4mol

4Al+3O2to→2Al2O34

3Fe+2O2to→Fe3O4

⇒nO2=\(\dfrac{3}{4}\)nAl+\(\dfrac{2}{3}\)nFe=\(\dfrac{5}{12}\)mol

⇒Vkk=\(\dfrac{5}{12}\).22,4.5=46,67l

b,

mrắn=27,8+mO2=27,8+​​\(\dfrac{5}{12}\)32=41,1g

→⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩Vkk=(0,15+415