a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)

PTHH : 3Fe + 2O₂ -t^o-> Fe₃O₄

              0,09    0,06        0,03

\(m_{Fe}=0,09.56=5,04\left(g\right)\)

\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)

b. PTHH : 2KCl + 3O₂ -> 2KClO₃

                            0,06           0,04

\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)

4

n _Fe3O4=\(\dfrac{6,96}{232}=0,03mol\)

3Fe+2O₂-to>Fe₃O₄

0,09---0,06-----0,03 mol

=>m _Fe=0,09.56=5,04g

=>V_O2=0,06.22,4=1,344l

b) 

2KClO₃-to>2KCl+3O₂

0,04----------------------0,06 mol

=>m _KClO3=0,04.122,5=4,9g

\(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=3.0,03=0,09\left(mol\right);n_{O_2}=0,03.2=0,06\left(mol\right)\\ a,\Rightarrow m_{Fe}=0,09.56=5,04\left(g\right);V_{O_2\left(đktc\right)}=0,06.22,4=1,344\left(l\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,06}{3}=0,04\left(mol\right)\\ \Rightarrow b,m_{KClO_3}=122,5.0,04=4,9\left(g\right)\)

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

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

a)\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)

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

0,4     \(\dfrac{4}{15}\)      \(\dfrac{2}{15}\)

\(V_{O_2}=\dfrac{4}{15}\cdot22,4=5,973l\)

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

   \(\dfrac{8}{45}\)                          \(\dfrac{4}{15}\)

  \(m_{KClO_3}=\dfrac{8}{45}\cdot122,5=21,78g\)

\(a/n_{Fe}=\dfrac{2,52}{56}=0,045mol\\ 3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ n_{O_2}=\dfrac{0,045.2}{3}=0,03mol\\ V_{O_2}=0,03.22,4=0,672l\\ b/2KClO_3\xrightarrow[]{t^0}2KCl+3O_2\\ n_{KClO_3}=\dfrac{0,03.2}{3}=0,02mol\\ m_{KClO_3}=0,02.122,5=2,45g\)

a)

PTHH: 3Fe + 2O₂  ^{____\(t^o\)____}> Fe₃O₄  (1) 

b) Ta có: n_Fe  = \(\dfrac{25.2}{56}=0.45\left(mol\right)\)

Theo (1): n_{\(O_2\)}= \(\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}0.45=0.3\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=0.3\cdot22.4=6.72\left(l\right)\)

c) PTHH: 2KClO₃ ^{__\(t^o\)___}>  2KCl + 3O₂   (2)

-Muốn điều chế được lượng oxi dùng cho phản ứng trên thì \(n_{O_2\left(2\right)}=n_{O_2\left(1\right)}=0.3\left(mol\right)\)

Theo (2) \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}0.3=0.2\left(mol\right)\)

=> \(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

Theo gt ta có: $n_{Mg}=0,15(mol)$

a, $2Mg+O_2\rightarrow 2MgO$

Ta có: $n_{O_2}=0,5.n_{Mg}=0,075(mol)\Rightarrow V_{O_2}=1,68(l)$

b, $2KClO_3\rightarrow 2KCl+3O_2$ (đk: nhiệt độ, MnO2)

Ta có: $n_{KClO_3}=\frac{2}{3}.n_{O_2}=0,05(mol)\Rightarrow m_{KClO_3}=6,125(g)$

\(n_{Mg}=\dfrac{3.6}{24}=0.15\left(mol\right)\)

\(2Mg+O_2\underrightarrow{t^0}2MgO\)

\(0.15......0.075......0.15\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(0.05.......................0.075\)

\(m_{KClO_3}=0.05\cdot122.5=6.125\left(g\right)\)

Theo gt ta có: nMg=0,15(mol)nMg=0,15(mol)

a, 2Mg+O2→2MgO2Mg+O2→2MgO

Ta có: nO2=0,5.nMg=0,075(mol)⇒VO2=1,68(l)nO2=0,5.nMg=0,075(mol)⇒VO2=1,68(l)

b, 2KClO3→2KCl+3O22KClO3→2KCl+3O2 (đk: nhiệt độ, MnO2)

Ta có: nKClO3=2/3.nO2=0,05(mol)⇒mKClO3=6,125(g)