\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

0,1         0,2    0,1

\(V_{O_2}=0,2\cdot22,4=4,48l\)

\(V_{CO_2}=0,1\cdot22,4=2,24l\)

a)

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

b) $n_{CH_4} = \dfrac{28}{22,4} = 1,25(mol)$

$n_{CO_2} = n_{CH_4} = 1,25(mol)$
$m_{CO_2} = 1,25.44 = 55(gam)$

c) $V_{O_2} =2 V_{CH_4} = 56(lít)$

a) PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\) \(\Rightarrow n_{O_2}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)

c) PTHH: \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)

Theo PTHH: \(n_{KClO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KClO_3}=0,1\cdot122,5=12,25\left(g\right)\)

a. \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)

PTHH : 3Fe + 2O₂ ---t^o---> Fe₃O₄  

             0,2      \(\dfrac{0.4}{3}\)

b. \(V_{O_2}=\dfrac{0.4}{3}.22,4=\dfrac{8.96}{3}\left(l\right)\)

c. PTHH : 2KClO₃ -> 2KCl + 3O₂   

                    \(\dfrac{0.8}{3}\)                     \(\dfrac{0.4}{3}\)

\(m_{KClO_3}=\dfrac{0.8}{3}.122,5=\dfrac{98}{3}\left(g\right)\)

3Fe+2O2-to>Fe3O4

0,45--0,3---------0,15 mol

n Fe=0,45 mol

=>VO2=0,3.22,4=6.72l

2KClO3-to>2KCl+3O2

0,2------------------------0,3

=>m KClO3=0,2.122,5=24,5g

THAM KHẢO:

Gọi số mol Mg và Zn lần lượt là x, y

Ta có 24x + 65y=23.3

     40x + 81y=36.1

=) x=0.7

    y= 0.1

b)

c)

\(a,2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ 2Zn+O_2\rightarrow\left(t^o\right)2ZnO\\ b,n_{O_2}=\dfrac{36,1-23,3}{32}=0,4\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,4.22,4=8,96\left(l\right)\\ \text{Đ}\text{ặt}:a=n_{Mg};b=n_{Zn}\left(a,b>0\right)\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}24a+65b=23,3\\0,5a+0,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,7\\b=0,1\end{matrix}\right.\\ c,m_{Mg}=24a=24.0,7=16,8\left(g\right)\\ m_{Zn}=65b=65.0,1=6,5\left(g\right)\)

a)

2Mg + O₂ --t^o--> 2MgO

2Zn + O₂ --t^o--> 2ZnO

b)

Gọi số mol Mg, Zn là a, b (mol)

=> 24a + 65b = 23,3 (1)

PTHH: 2Mg + O₂ --t^o--> 2MgO

               a-->0,5a------>a

            2Zn + O₂ --t^o--> 2ZnO

               b-->0,5b------>b

=> 40a + 81b = 36,1 (2)

(1)(2) => a = 0,7 (mol); b = 0,1 (mol)

\(n_{O_2}=0,5a+0,5b=0,4\left(mol\right)\)

=> \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

c) 

m_Mg = 0,7.24 = 16,8 (g)

m_Zn = 0,1.65 = 6,5 (g)

Câu 3.

a)\(n_{Fe}=\dfrac{12,6}{56}=0,225mol\)

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

0,225  0,15    0,075

\(V_{O_2}=0,15\cdot22,4=3,36l\)

b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

   0,1                            0,15

\(m_{KClO_3}=0,1\cdot122,5=12,25g\)

c)\(n_S=\dfrac{12,8}{32}=0,4mol\)

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

0,4   0,15   0

0,15 0,15   0,15

0,25 0        0,15

\(m_{Sdư}=0,25\cdot32=8g\)