nH2 = 13,44/22,4 = 0,6 (mol)

PTHH: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

Mol: 0,4 <--- 0,6 <--- 0,2 <--- 0,6

mAl = 0,4 . 27 = 10,8 (g)

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

PTHH: 2H2 + O2 -> (t°) 2H2O

LTL: 0,6/2 < 0,5 => O2 dư

VO2 (p/ư) = 13,44/2 = 6,72 (l)

VO2 (dư) = 11,2 - 6,72 = 4,48 (l)

nH2O = 0,6 (mol)

mH2O = 0,6 . 18 = 10,8 (g)

a.\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,4                                             0,6    ( mol )

\(m_{Al}=0,4.27=10,8g\)

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

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,6 <   0,5                         ( mol )

0,6      0,3             0,6           ( mol )

\(V_{O_2\left(dư\right)}=\left(0,5-0,3\right).22,4=4,48l\)

\(m_{H_2O}=0,6.18=10,8g\)

nH2 = \(\dfrac{13,44}{22,4}\)= 0,6 (mol)

PTHH: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

Mol: 0,4 <--- 0,6 <--- 0,2 <--- 0,6

mAl = 0,4 . 27 = 10,8 (g)

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

PTHH: 2H2 + O2 -to> 2H2O

=> O2 dư

VO2 pứ = 13,44/2 = 6,72 (l)

VO2 dư = 11,2 - 6,72 = 4,48 (l)

nH2O = 0,6 (mol)

mH2O = 0,6 . 18 = 10,8 (g)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

a/ \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Lập tỉ lệ :

\(\dfrac{n_{Zn}}{2}< n_{O_2}\)

=> \(O_2\) dư , Zn phản ứng hết

b/ \(n_{ZnO}=n_{Zn}=0,2\left(mol\right)\)

=> \(m_{ZnO}=16,2\left(g\right)\)

c/ Cần bổ sung Zn để phản ứng xảy ra hoàn toàn

\(n_{O_2}=\dfrac{V}{24,79}=\dfrac{5,6}{24,79}\approx0,23\left(mol\right)\\ n_P=\dfrac{m}{M}=\dfrac{3,1}{31}=0,1\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\) 

                4        5          2

              0,1    0,125     0,05

a. Tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,12}{5}\Rightarrow O_2\) dư và dư \(0,025-0,024=0,001\left(mol\right)\\ m_{O_2}=n.M=0,001.\left(16.2\right)=0,032\left(g\right)\)  

b. \(m_{P_2O_5}=n.M=0,05.\left(31.2+16.5\right)=7,1\left(g\right).\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ Vì:\dfrac{0,1}{4}< \dfrac{1}{5}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=1-\dfrac{5}{4}.0,1=0,875\left(mol\right)\\ m_{O_2}=0,875.32=28\left(g\right)\\ n_{P_2O_5}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right);n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o→ 2P₂O₅

Mol:     0,1    0,125          0,05

Ta có: \(\dfrac{0,1}{4}< \dfrac{1}{5}\) ⇒ P hết, O₂ dư

\(m_{O_2dư}=\left(1-0,125\right).32=28\left(g\right)\)

\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

Ta có : 

\(n_{Al_2O_3}=\dfrac{0.2\cdot2}{4}=0.1\left(mol\right)\)

\(m_{Al_2O_3}=0.1\cdot102=10.2\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2.......\dfrac{2}{15}.....\dfrac{1}{15}\)

\(V_{O_2}=\dfrac{2}{15}\cdot22.4=2.98\left(l\right)\)

\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.46\left(g\right)\)

\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)

Ta có :

\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)

a/ Ta có: \(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

PTHH:

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

4       5

x       0.2

\(=>x=\dfrac{4\cdot0.2}{5}=0.16=n_P\)

\(=>m_P=0.16\cdot31=4.96\left(g\right)\) hay a=4.96

b/ PTHH:

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

4                  2

0.16             y

\(=>y=\dfrac{0.16\cdot2}{4}=0.08=n_{P_2O_5}\)   

\(=>m_{P_2O_5}=0.08\cdot\left(31\cdot2+16\cdot5\right)=11.36\left(g\right)\)

a/ Ta có: \(n_S=\dfrac{0.32}{32}=0.01\left(mol\right)\)

PTHH: 

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

1     1

0.01 x

\(=>x=\dfrac{0.01\cdot1}{1}=0.01=n_{O_2}\)

\(=>V_{O_2}=0.01\cdot22.4=0.224\left(l\right)\)

b/ PTHH:

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

1              1

0.01         y

\(=>y=\dfrac{0.01\cdot1}{1}=0.01=n_{SO_2}\)

\(=>m_{SO_2}=0.01\cdot\left(32+16\cdot2\right)=0.64\left(g\right)\)