a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)$
$V_{O_2} = 0,15.22,4 = 3,36(lít)$

b) $2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2$

$n_{KClO_3\ pư} = \dfrac{2}{3}n_{O_2} = 0,1(mol)$
$m_{KClO_3\ pư} = 0,1.122,5 = 12,25(gam)$

$\Rightarrow m_{KClO_3\ đã\ dùng} = 12,25 : (100\% - 10\%) = 13,61(gam)$

$n_X = \dfrac{2,8}{22,4} = 0,125(mol)$

$n_{CO_2} = \dfrac{8,4}{22,4} = 0,375(mol)$
$n_{H_2O} = \dfrac{6,75}{18} = 0,375(mol)$

Bảo toàn nguyên tố C, H : 

$n_C = n_{CO_2} = 0,375(mol) ; n_H = 2n_{H_2O} = 0,75(mol)$
Suy ra : 

Số nguyên tử C = $\dfrac{0,375}{0,125} = 3$

Số nguyên tử H = $\dfrac{0,75}{0,125} = 6$

Vậy CTPT là $C_3H_6$

CTCT : $CH_2=CH-CH_3$

1) 2-metylbutan

2) Từ \(CH_3-CH_2-CH\left(CH_3\right)-CH\left(C_2H_5\right)-CH_3\) ta bẻ mạch thành

\(CH_3-CH_2-CH\left(CH_3\right)-CH\left(CH_3\right)-CH_2-CH_3\)

2,3-đimetylhexan

3) 2,2-đimetylbutan

4) Từ \(CH_3-CHBr-CH_2-CH\left(C_2H_5\right)-CH_3\) bẻ mạch ta được:

\(CH_3-CHBr-CH_2-CH\left(CH_3\right)-CH_2-CH_3\)

2-brom-4-metylhexan

- C₅H₁₂:

\(CH_3-CH_2-CH_2-CH_2-CH_3:\) n-pentan

\(CH_3-CH\left(CH_3\right)-CH_2-CH_3:\) 2-metylbutan

\(CH_3-C\left(CH_3\right)_2-CH_3:\) 2,2-đimetylpropan

- C₆H₁₄:

\(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3:\) n-hexan

\(CH_3-CH\left(CH_3\right)-CH_2-CH_2-CH_3:\) 2-metylpentan

\(CH_3-CH_2-CH\left(CH_3\right)-CH_2-CH_3:\) 3-metylpentan

\(CH_3-CH\left(CH_3\right)-CH\left(CH_3\right)-CH_3:\) 2,3-đimetylbutan

- C₇H₁₆:

\(CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3:\) n-heptan

\(CH_3-CH\left(CH_3\right)-CH_2-CH_2-CH_2-CH_3:\) 2-metylhexan

\(CH_3-CH_2-CH\left(CH_3\right)-CH_2-CH_2-CH_3:\) 3-metylhexan

\(CH_3-CH\left(CH_3\right)-CH\left(CH_3\right)-CH_2-CH_3:\) 2,3-đimetylpentan

\(CH_3-CH\left(CH_3\right)-CH_2-CH\left(CH_3\right)-CH_3:\) 2,4-đimetylpentan

\(CH_3-C\left(CH_3\right)_2-CH_2-CH_2-CH_3:\) 2,2-đimetylpentan

\(CH_3-CH_2-C\left(CH_3\right)_2-CH_2-CH_3:\) 3,3-đimetylpentan

\(CH_3-CH_2-CH\left(C_2H_5\right)-CH_2-CH_3:\) 3-etylpentan

\(CH_3-C\left(CH_3\right)_2-CH\left(CH_3\right)-CH_3:\) 2,2,3-trimetylbutan

Khí B gồm : $CO(a\ mol) ; CO_2(b\ mol)$

Ta có : 

$a + b = \dfrac{11,2}{22,4} = 0,5(mol)$
$M_B = \dfrac{28a + 44b}{a + b} = 20,4.2$

Suy ra : a = 0,1 ; b = 0,4

$CO + O_{oxit} \to CO_2$
$n_{O(oxit)\ pư} = n_{CO_2} = 0,4(mol)$
$\Rightarrow m = m_A + m_O = 64 + 0,4.16 = 70,4(gam)$

a) $n_{Al} = \dfrac{0,81}{27}  = 0,03(mol) ; n_{HCl} = \dfrac{1,825}{36,5} = 0,05(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$

Vì : 

$n_{Al} : 2 > n_{HCl} : 6$ nên Al dư

$n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,025(mol)$

$V_{H_2} = 0,025.24,79 = 0,61975(lít)$

b) $n_{Al\ pư} = \dfrac{1}{3} n_{HCl} = \dfrac{0,05}{3}(mol)$
Ta thấy : $m_{Al} - m_{H_2} = \dfrac{0,05}{3}.27 - 0,025.2 = 0,4 > 0$
Do đó, dung dịch tăng so với khối lượng dung dịch HCl ban đầu 0,4 gam

1)

$CH_4 +2 O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

Theo PTHH : 

$V_{O_2\ cần\ dùng} = 2V_{CH_4} = 24,79(lít)$
$V_{CO_2} = V_{CH_4} = 12,395(lít)$

2)

a)

$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
$V_{O_2} = 3V_{C_2H_4} = 14,874(lít)$

b) $V_{không\ khí} = V_{O_2} : 20\% = 14,874 : 20\% = 74,37(lít)$

n_Zn = m_Zn:M_Zn = 65:65 = 1mol

n_HCl = m_HCl:M_HCl = 36,5:36,5 = 1mol

          Zn + 2HCl -> ZnCl₂ + H₂

Theo PTHH:1mol 2mol            1mol

Theo ĐB:  1mol 1mol            1mol

a, Vì 1mol/1mol > 1mol/2mol => HCl sẽ hết trước => Tính theo Zn

b, m_H2 = n_H2.22,4 = 2.22,4 = 44,8ml

\(n_{HCl}=\dfrac{m}{M}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{65}{65}=1\left(mol\right)\)

                     Zn  +   2HCl   →  ZnCl₂ + H₂

Theo đề:    1 mol     2 mol

Pư:             1 mol    1 mol

Tỉ lệ:           \(\dfrac{1}{1}>\dfrac{1}{2}\)             \(\Rightarrow\) Zn là chất thừa, tính theo HCl

   Zn  +   2HCl   →  ZnCl₂ + H₂

                1                   →    1             ( mol )

\(V_{H_2}=n.22,4=1.22,4=22,4\left(l\right)\)