C+O2-to>CO2

Áp dụng định luật bảo toàn khối lượng

m cacbon + m khí oxi = m khí cacbonic

=> m khí cacbonic = 4,5 + 12 = 16,5kg

Gọi CTHH là \(C_xH_y\)

\(n_{CO_2}=\dfrac{8,8}{44}=0,2mol\Rightarrow m_C=2,4g\)

\(d_A\)_/O2=0,5\(\Rightarrow M_A=0,5\cdot32=16đvC\)

\(n_A=\dfrac{3,2}{16}=0,2mol\)

\(\overline{C}=\dfrac{n_{CO_2}}{n_A}=\dfrac{0,2}{0,2}=1\)

Vậy A là \(CH_4\).

a.\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2mol\)

\(n_{O_2}=\dfrac{m}{M}=\dfrac{9,6}{32}=0,3mol\)

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

Xét: \(\dfrac{0,2}{4}\) < \(\dfrac{0,3}{5}\)                         ( mol )

      0,2                          0,1             ( mol )

\(m_{P_2O_5}=n.M=0,1.142=14,2g\)

b.\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{17,4}{232}=0,075mol\)

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

          0,15                 0,075   ( mol )

\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

   0,1                                               0,15  ( mol )

\(m_{KClO_3}=n.M=0,1.122,5=12,25g\)

Tính khối lượng:

a) 0,3 mol khí CO₂

(CO₂ = 12 + 16.2 = 44 g/mol)

=. m = n . M = 0,3 . 44 = 13,2g

b) 67,2 lít khí CH₄ (ở đktc)

n = \(\dfrac{V}{24,79}=\dfrac{67,2}{24,79}=2,710mol\)  

(CH₄ = 12 + 1.4 = 16 g/mol)

=> m = n . M = 2,710 . 16 = 43,36g

c) 896 cm³ khí CO₂ 

896 cm³ = 0,896l

\(n=\dfrac{V}{24,79}=\dfrac{0,896}{24,79}=0,036mol\)

(CO₂ = 12 + 16.2 = 44 g/mol)

=> m = n . M = 0,036 . 44 = 1,584g

d) 11,2 dm³ khí O₂

11,2 dm³ = 0,0112l

\(n=\dfrac{V}{24,79}=\dfrac{0,0112}{24,79}=4,517mol\) 

(O₂ = 16.2 = 32 g/mol)

=> m = n . M = 4,517 . 32 = 144,544g.

Coi $n_A = 1(mol) \Rightarrow m_A = 1.2.13,5 = 27(gam)$
$m_{NH_3} + m_{O_2} + m_{N_2} = 27$

$\Rightarrow \dfrac{7}{8}m_{O_2} + m_{O_2} + \dfrac{3}{6} (m_{O_2} + m_{NH_3} ) = 27$

$\Rightarrow \dfrac{7}{8}m_{O_2} + m_{O_2} + \dfrac{3}{6} (m_{O_2} + \dfrac{7}{8}m_{O_2} ) = 27$
$\Rightarrow \dfrac{45}{16}m_{O_2} = 27 \Rightarrow m_{O_2} = 9,6(gam)$

Suy ra: 

$m_{NH_3} = 8,4 ; m_{N_2} = 9$

Suy ra : $n_{O_2} = 0,3(mol) ; n_{NH_3} = \dfrac{42}{85}(mol)$

$\%V_{O_2} = \dfrac{0,3}{1}.100\% = 30\%$
$\%V_{NH_3} = 49,41\%$
$\%V_{N_2} = 20,59\%$

nO2 = 1,12/22,4 = 0,05 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

Mol: 0,04 <--- 0,05

mP = 0,04 . 31 = 1,24 (g)

Có: \(\%O=\dfrac{16y}{M_A.x+16y}.100\%=50\%\)

=> M_A.x = 16y

Giả sử \(m_{O_2}=m_{A_xO_y}=32\left(g\right)\)

\(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\Rightarrow n_{A_xO_y}=\dfrac{1}{2}=0,5\left(mol\right)\)

=> \(M_{A_xO_y}=\dfrac{32}{0,5}=64\left(g/mol\right)\)

=> M_A.x + 16y = 64

=> 32y = 64 

=> y = 2

=> M_A.x = 32

Chỉ có x = 1 thỏa mãn => M_A = 32 (g/mol)

=> A là S

CTHH: SO₂