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

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

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

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

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

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

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

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

a)

\(n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,075(mol)\\ \Rightarrow V_{O_2} = 0,075.22,4 = 1,68(lít)\)

b)

\(2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{2}{3}.0,075 = 0,05(mol)\\ \Rightarrow m_{KClO_3} = 0,05.122,5 = 6,125(gam)\)

mMg = 3.6/24 = 0.15 (mol) 

2Mg + O2 -to-> 2MgO 

0.15__0.075____0.15 

mMgO= 0.15*40 = 6 (g) 

VO2  = 0.075*22.4 = 1.68 (l) 

2KClO3 -to-> 2KCl + 3O2 

0.05_______________0.075 

mKClO3 = 0.05*122.5 = 6.125 (g) 

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_{O_2} = \dfrac{5}{4}n_P = \dfrac{5}{4}.\dfrac{3,72}{31} = 0,15(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)\\ \Rightarrow m_{KClO_3} = 0,1.122,5 = 12,25(gam)\)

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

a.\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

0,1       0,05                       ( mol )

\(V_{kk}=\left(0,05.22,4\right).5=5,6l\)

b.\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

      1/30                                0,05      ( mol )

\(m_{KClO_3}=\dfrac{1}{30}.122,5=4,08g\)

2KClO₃-to\xt->2KCl+3O₂

0,1------------------0,1

n KClO3=\(\dfrac{12,25}{122,5}\)=0,1 mol

=>m KCl=0,1.74,5=7,45g

H=\(\dfrac{6,8}{7,45}.100\)=91,275%

b)

2KClO₃-to\xt->2KCl+3O₂

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

n O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

H=85%

=>m KClO3=0,2.122,5.\(\dfrac{100}{85}\)=28,82g

c)

2KClO₃-to\xt->2KCl+3O₂

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

n KClO3=\(\dfrac{24,5}{122,5}\)=0,2 mol

H=80%

=>m O2=0,3.32.\(\dfrac{80}{100}\)=10,4g