Bài 9

PTHH: $2KClO_3 \overset{t^{\circ}}{\rightarrow} 2KCl + 3O_2$

$n_{KClO_3}=$ $\dfrac{24,5}{122,5}=0,2(mol)$

Theo PTHH $\to$ $n_{O_2}=$ $\dfrac{0,2*3}{2}=0,3(mol)$ 

$\to$ $V_{O_2}=0,3*22,4=6,72(l)$

Bài 10

Đổi $784ml=0,784(l)$

PTHH $2Mg + O_2 \overset{t^{\circ}}{\rightarrow} 2MgO$

$n_{Mg}=$ $\dfrac{2,4}{24}=0,1(mol)$

$n_{O_2}=$ $\dfrac{0,784}{22,4}=0,035(mol)$

Có $\dfrac{0,1}{2}>0,035\to $ Sau phản ứng $O_2$ hết, $Mg$ dư

$→m_{O_2}=0,035*32=1,12(g)$

Bảo toàn khối lượng $→m_{Mg}+m_{O_2}=m_{rắn}$

$→m_{rắn}=1,12+2,4=3,52(g)$

Bài 11:

$n_{O_2}=$ $\dfrac{3,36}{22,4}=0,15(mol)$ 

$n_{Al}=$ $\dfrac{2,7}{27}=0,1(mol)$

PTHH $2Mg + O_2 \overset{t^{\circ}}{\rightarrow} 2MgO(1)$

$4Al + 3O_2 \overset{t^{\circ}}{\rightarrow} 2Al_2O_3(2)$

Theo PTHH $(2)→$ $n_{O_2}(2)=$ $\dfrac{0,1*3}{4}=0,075(mol)$ 

$→n_{O_2}(1)=0,15-0,075=0,075(mol)$

Theo PTHH $(1) →$ $n_{Mg}=$ $\dfrac{0,075*2}{1}=0,15(mol)$ 

$→ m_{Mg}=0,15*24=3,6(g)$

$→$ $\%m_{Mg}=$ $\dfrac{3,6}{3,6+2,7}*100\%=57,14\%$

$→$ $\%m_{Al}=100\%-57,14\%=42,86\%$

9.PTHH : 2KClO3 -> 2KCl + 3O2
nKClO3 : 24,5/122,5 = 0,2 mol

-> nO2 = 0,2*3/2 = 0,3 mol
-> VO2 = 0,3*22,4 = 6,72
10. PTHH : 2Mg + O2 -> 2MgO
nMg : 2,4/24 = 0,1 mol
nO2 : 0,784/22,4 = 0,035 mol
ta có nMg/2 > nO2/1 nên Mg dư
-> tính theo O2
-> nMgO = 0,035*2/1 = 0,07 mol
-> mMgO = 0,07*24=1,68(g)