nO2=6,72/22,4=0,3(mol)

PTHH: 2 KClO3 -to->2  KCl +3 O2

Ta có: nKClO3=2/3. 0,3=0,2(mol)

=>mKClO3=0,2.122,5=24,5(g)

đây là hóa học ak bạn

 a) n_O2 = 4,48 / 22,4 = 0,2 (mol)

             2KClO₃  -------t^o-------> 2KCl + 3O₂

m _KClO3 = 122,5 * ( 0,2 * 2 / 3 ) =49/3 g

b) mol KCl thu đc

 n_KCl = 1, 5*2 /2 = 1,5 mol

m_KCl = 1,5 *74,5 =111, 75 g

PT: 2KClO₃ ------t*----- > 2KCl + 3O₂

a, Số mol O₂ cần điều chế là: \(n_{O_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT,ta có: \(n_{KClO_3}=\frac{2}{3}.n_{O_2}=\frac{2}{3}.0,2=\frac{2}{15}\left(mol\right)\)

Khối lượng KClO₃ cần dùng là: \(m_{KClO_3}=\frac{2}{15}.122,5=\frac{49}{3}\approx16,33\left(g\right)\)

b,Theo PT + đb ,ta có: \(n_{KCl}=n_{KClO_3}=1,5\left(mol\right)\)

=> Khối lượng KCl thu đc là: \(m_{KCl}=1,5.74,5=111,75\left(g\right)\)

=.= hok tốt!!

\(a.n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow n_{KClO_3}=0,2.\dfrac{2}{3}=\dfrac{2}{15}\left(mol\right)\\ m_{KClO_3}=\dfrac{2}{15}.122,5\approx16,333\left(g\right)\\ b.n_{KClO_3}=1,5\left(mol\right)\Rightarrow n_{O_2}=\dfrac{3}{2}.1,5=2,25\left(mol\right)\\ m_{O_2}=2,25.32=144\left(g\right)\\ c.n_{KClO_3}=0,1\left(mol\right)\\ \Rightarrow n_{KCl}=n_{KClO_3}=0,1\left(mol\right);n_{O_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)

\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)

\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)

_________0,1_____________0,15______mol

\(\Rightarrow m_{KClO3}=0,1.122,5=12,25\left(g\right)\)

\(2KClO_3\rightarrow2KCl+3O_2\)

\(n_{O_2}=\frac{3.36}{22.4}=0.15\left(mol\right)\\ \Rightarrow n_{KClO_3}=\frac{2}{3}\cdot n_{O_2}=\frac{2}{3}\cdot0.15=0.1\left(mol\right)\\ \Rightarrow m_{KClO_3}=0.1\cdot122.5=12.25\left(g\right)\)

a)2KClO3−−>2KCl+3O2

0,1---------------0,1--------0,15 mol

nO2=3,36\22,4=0,15(mol)

mKClO3=0,1.122,5=12,25(g)

2Mg+O₂-to>2MgO

0,15---0,075 mol

2KClO₃-to->2KCl+3O₂

0,05-----------------------0,075

n _Mg=\(\dfrac{3,6}{24}\)=0,15 mol

=>V_O2=0,075.22,4=1,68l

=>m _KClO3=0,05.122,5=6,125g

Sao lại cho Mg vào mà ''khối lượng kali clorat'' là sao em

 a. \(n_{Mg}=\dfrac{3.6}{24}=0,15\left(mol\right)\)

PTHH : 2Mg + O₂ ----t^o----> 2MgO

             0,15   0,075             

\(V_{O_2}=0,075.22,4=1,68\left(l\right)\)

b. PTHH : 2KClO₃ \(\xrightarrow[MnO_2]{t^o}\) 2KCl + 3O₂

                   0,05                         0,075

\(m_{KClO_3}=0,05.122,5=6,125\left(g\right)\)

Câu 6.

\(n_{O_2}=\dfrac{16,8}{22,4}=0,75mol\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

1,5                                                 0,75

\(m_{KMnO_4}=1,5\cdot158=237g\)

Câu 7.

\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)

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

           0,04      0,02

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

\(\dfrac{2}{75}\)                           0,04

\(m_{KClO_3}=\dfrac{2}{75}\cdot122,5=\dfrac{49}{15}\approx3,27g\)

Bài 6 :

\(n_{O_2}=\dfrac{16.8}{22,4}=0,75\left(mol\right)\)

PTHH : 2KMnO₄ ----t⁰-----> K₂MnO₄ + MnO₂ + O₂

               1,5                                                        0,75

\(m_{KMnO_4}=1,5.158=237\left(g\right)\)

\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

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

Theo PT: \(n_{KMnO_4}=2n_{O_2}=1\left(mol\right)\Rightarrow m_{KMnO_4}=1.158=158\left(g\right)\)

\(n_{CH_4} = \dfrac{5,6}{22,4} = 0,25(mol)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ n_{H_2O} = n_{O_2} = 2n_{CH_4} = 0,5(mol)\\ \Rightarrow V_{O_2} = 0,5.22,4 = 11,2(lít) ; m_{H_2O} = 0,5.18 = 9(gam);\\ V_{kk} = 5V_{O_2} = 11,2.5 = 56(lít)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ m_{KMnO_4} = 0,5.2.158 = 158(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl +3O_2\\ m_{KClO_3} = 0,5.\dfrac{2}{3}.122,5 = 40,83\ gam\)