a)

$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$S + O_2 \xrightarrow{t^o} SO_2$

b) Gọi $n_P = a(mol) ; n_S = b(mol) \Rightarrow 31a + 32b = 15,6(1)$

Theo PTHH : 

$n_{O_2} = \dfrac{5}{4}n_P + n_S = \dfrac{5a}{4} + b = \dfrac{13,44}{22,4} = 0,6(2)$

Từ (1)(2) suy ra : a = 0,4 ; b = 0,1

$m_P = 0,4.31 = 12,4(gam)$
$m_S = 0,1.32 = 3,2(gam)$

c) $n_{P_2O_5} = \dfrac{1}{2}n_P = 0,2(mol) \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)$
$n_{SO_2} = n_S = 0,1(mol) \Rightarrow V_{SO_2} = 0,1.22,4 = 2,24(lít)$

a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

                \(4H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)

a)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,1<-0,05<-------0,1

            2CO + O₂ --t^o--> 2CO₂

            0,2<--0,1-------->0,2

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)

\(2CO+O_2\rightarrow2CO_2\)

  a         0,5a     a

\(2H_2+O_2\rightarrow2H_2O\)

 0,1   0,05 \(\leftarrow\)    0,1

\(\Sigma n_{O_2}=0,5a+0,05=0,15\)

\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)

\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{CO_2}=0,2\cdot44=8,8g\)

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

a. PTHH:

Cu + H₂SO₄ ---x--->

Mg + H₂SO₄ ---> MgSO₄ + H₂

b. Ta có: \(n_{H_2SO_4}=2.\dfrac{100}{1000}=0,2\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)

\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{6}.100\%=80\%\)

\(\%_{m_{Cu}}=100\%-80\%=20\%\)

c. Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

d. PTHH: \(Cu+2H_2SO_{4_đ}\overset{t^o}{--->}CuSO_4+SO_2+2H_2O\)

a) \(n_{SO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

         0,5<-0,5<------0,5

=> m_S = 0,5.32 = 16(g)

=> \(\left\{{}\begin{matrix}\%m_S=\dfrac{16}{22,2}.100\%=72,07\%\\\%m_P=\dfrac{22,2-16}{22,2}.100\%=27,93\%\end{matrix}\right.\)

b) \(n_P=\dfrac{22,2-16}{31}=0,2\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

          0,2-->0,25----->0,1

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

c) 

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

             0,5<-------------------0,75

=> \(m_{KClO_3}=0,5.122,5=61,25\left(g\right)\)

a) PTHH:

\(S+O_2\rightarrow\left(t^o\right)SO_2\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

- Chất khí mùi hắc là SO₂

- Chất rắn sau phản ứng có m(g) là P₂O₅

Đặt: n_S=a(mol); n_P=b(mol) (a,b>0) (nguyên, dương)

\(\Rightarrow\left\{{}\begin{matrix}32a+31b=22,2\\22,4a=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_S=\dfrac{0,5.32}{22,2}.100\approx72,072\%\\\%m_P\approx100\%-72,072\%\approx27,928\%\end{matrix}\right.\)

b)

\(n_{O_2}=a+\dfrac{5}{4}b=0,5+\dfrac{5}{4}.0,2=0,75\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,75.22,4=16,8\left(l\right)\)

c)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,75}{3}=0,5\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.0,5=61,25\left(g\right)\)

\(n_{CO_2}=\dfrac{8.8}{44}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)

\(2CO+O_2\underrightarrow{t^0}2CO_2\)

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

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.4......0.3-0.1\)

\(\%m_{CO}=\dfrac{0.2\cdot28}{0.2\cdot28+0.4\cdot2}\cdot100\%=87.5\%\)

\(\%m_{H_2}=100-87.5=12.5\%\)

a)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2) \)

b)

\(n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\)

Theo PTHH :

\(n_{CO} = n_{CO_2} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{1}{2}n_{CO_2} = 0,1(mol)\\ n_{H_2} = 2n_{O_2(2)} = 2(0,3-0,1) = 0,4(mol)\)

Vậy :

\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

a/ \(2CO\left(0,2\right)+O_2\left(0,1\right)\rightarrow2CO_2\left(0,2\right)\)

\(2H_2\left(0,1\right)+O_2\left(0,05\right)\rightarrow2H_2O\left(0,1\right)\)

\(n_{H_2O}=\frac{1,8}{18}=0,1\)

\(n_{O_2}=\frac{3,36}{22,4}=0,15\)

Số mol O₂ phản ứng ở phản ứng đầu là: \(0,15-0,05=0,1\)

\(\Rightarrow m_{CO_2}=0,2.44=8,8\)

b/ \(m_{CO}=0,2.28=5,6\)

\(m_{H_2}=0,1.2=0,2\)

c/ \(\%CO=\frac{0,2}{0,3}.100\%=66,67\%\)

\(\Rightarrow\%H_2=100\%-66,67\%=33,33\%\)

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

              a___a______a    (mol)

            \(S+O_2\xrightarrow[]{t^o}SO_2\)

             b___b_______b   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}12a+32b=10\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_C=0,3\cdot12=3,6\left(g\right)\\m_S=6,4\left(g\right)\\V_{khí}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)