Đáp án + Giải thích các bước giải:

Câu 26:

$1)S+O_2\xrightarrow{t^o}SO_2$

$2)4P+5O_2\xrightarrow{t^o}2P_2O_5$

$3)C+O_2\xrightarrow{t^o}CO_2$

$4)3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$5)4Na+O_2\xrightarrow{t^o}2Na_2O$

$6)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$7)2H_2+O_2\xrightarrow{t^o}2H_2O$

$8)H_2+CuO\xrightarrow{t^o}Cu+H_2O$

$9)3H_2+Fe_2O_3\xrightarrow{t^o}2Fe+3H_2O$

$10)BaO+H_2O\to Ba(OH)_2$

$11)SO_2+H_2O\to H_2SO_3$

$12)Zn+2HCl\to ZnCl_2+H_2$

$13)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$14)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$15)2KClO_3\xrightarrow{t^o}2KCl+3O_2$

$16)2Na+2H_2O\to 2NaOH+H_2$

Câu 27:

$a)2KClO_3\xrightarrow{t^o}2KCl+3O_2$

Theo PT: $n_{O_2}=1,5n_{KClO_3}=1,5.\frac{24,5}{122,4}=0,3(mol)$

$\to V_{O_2}=0,3.22,4=6,72(l)$

$b)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

Theo PT: $n_{KMnO_4}=2n_{O_2}=2.\frac{5,6}{22,4}=0,5(mol)$

$\to m_{KMnO_4}=0,5.158=79(g)$

Câu 28:

$a)n_{Fe_3O_4}=\frac{23,2}{232}=0,1(mol)$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$n_{Fe}=0,3(mol)\to m_{Fe}=0,3.56=16,8(g)$

$b)n_{O_2}=0,2(mol)\to V_{O_2}=0,2.22,4=4,48(l)$

$c)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$n_{KMnO_4}=0,4(mol)\to m_{KMnO_4}=0,4.158=63,2(g)$

Câu 29:

$a)2Al+6HCl\to 2AlCl_3+3H_2$

$b)AlCl_3:$ nhôm clorua, $H_2:$ hiđro

$n_{Al}=\frac{8,1}{27}=0,3(mol)$

$n_{AlCl_3}=0,3(mol);n_{H_2}=0,45(mol)$

$m_{AlCl_3}=0,3.133,5=40,05(g);m_{H_2}=0,45.2=0,9(g)$

Câu 30:

$a)n_{Na}=\frac{13,8}{23}=0,6(mol)$

$2Na+2H_2O\to 2NaOH+H_2$

$n_{NaOH}=0,6(mol)\to m_{NaOH}=0,6.40=24(g)$

$n_{H_2}=0,3(mol)\to V_{H_2}=0,3.22,4=6,72(l)$

$b)n_{Fe_2O_3}=\frac{19,2}{160}=0,12(mol)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$n_{Fe_2O_3}>\frac{n_{H_2}}{3}\to F_2O_3$ dư

$n_{Fe}=0,2(mol)\to m_{Fe}=0,2.56=11,2(g)$

Đáp án:

Câu 26:

1. $S + O_{2}\xrightarrow{t^o}SO_{2}$

2. $4P + 5O_{2}\xrightarrow{t^o}2P_{2}O_{5}$

3. $C + O_{2}\xrightarrow{t^o}CO_{2}$

4. $3Fe + 2O_{2}\xrightarrow{t^o}Fe_{3}O_{4}$

5. $4Na + O_{2}\xrightarrow{t^o}2Na_{2}O$

6. $4Al + 3O_{2}\xrightarrow{t^o}2Al_{2}O_{3}$

7. $2H_{2} + O_{2}\xrightarrow{t^o}2H_{2}O$

8. $CuO + H_{2}\xrightarrow{t^o}Cu + H_{2}O$

9. $Fe_{2}O_{3} + 3H_{2}\xrightarrow{t^o}2Fe + 3H_{2}O$

10. $BaO + H_{2}O→Ba(OH)_{2}$

11. $SO_{2} + H_{2}O→H_{2}SO_{3}$

12. $Zn + 2HCl→ZnCl_{2} + H_{2}↑$

13. $2Al + 3H_{2}SO_{4}→Al_{2}(SO_{4})_{3} + 3H_{2}↑$

14. $2KMnO_{4}\xrightarrow{t^o}K_{2}MnO_{4} + MnO_{2}↓ + O_{2}↑$

15. $2KClO_{3}\xrightarrow{t^o}2KCl + 3O_{2}↑$

16. $2Na + 2H_{2}O→2NaOH + H_{2}↑$

Câu 27:

a) $2KClO_{3}\xrightarrow{t^o}2KCl + 3O_{2}↑$

Ta có: $n_{KClO_{3}}=\dfrac{m}{M}=\dfrac{24,5}{122,5}= 0,2 (mol)$

Theo PT: $n_{O_{2}}=\dfrac{3}{2}n_{KClO_{3}}=\dfrac{3}{2} × 0,2= 0,3 (mol)$

⇒ $V_{O_{2}}=n × 22,4=0,3 × 22,4= 6,72 (l)$

b) $2KMnO_{4}\xrightarrow{t^o}K_{2}MnO_{4} + MnO_{2}↓ + O_{2}↑$

Ta có: $n_{O_{2}}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}= 0,25 (mol)$

Theo PT: $n_{KMnO_{4}}=2n_{O_{2}}=2 × 0,25= 0,5 (mol)$

⇒ $m_{KMnO_{4}}=n × M=0,5 × 158= 79 (g)$

Câu 28:

$3Fe + 2O_{2}\xrightarrow{t^o}Fe_{3}O_{4}$ (1)

$2KMnO_{4}\xrightarrow{t^o}K_{2}MnO_{4} + MnO_{2}↓ + O_{2}↑$ (2)

a) Ta có: $n_{Fe_{3}O_{4}}=\dfrac{m}{M}=\dfrac{2,32}{232}= 0,01 (mol)$

Theo PT (1): $n_{Fe}=3n_{Fe_{3}O_{4}}=3 × 0,01= 0,03 (mol)$

⇒ $m_{Fe}=n × M=0,03 × 56= 1,68 (g)$

b) Theo PT (1): $n_{O_{2}}=2n_{Fe_{3}O_{4}}=2 × 0,01= 0,02 (mol)$

⇒ $V_{O_{2}}=n × 22,4=0,02 × 22,4= 0,448 (l)$

c) Theo PT (2): $n_{KMnO_{4}}=2n_{O_{2}}=2 × 0,02= 0,04 (mol)$

⇒ $m_{KMnO_{4}}=n × M=0,04 × 158= 6,32 (g)$

Câu 29:

a) $2Al + 6HCl→2AlCl_{3} + 3H_{2}↑$

b) $AlCl_{3}$: Nhôm clorua

$H_{2}$: Khí hiđro

Ta có: $n_{Al}=\dfrac{m}{M}=\dfrac{8,1}{27}= 0,3 (mol)$

Theo PT: $n_{AlCl_{3}}=n_{Al}= 0,3 (mol)$

Theo PT: $n_{H_{2}}=\dfrac{3}{2}n_{H_{2}}=\dfrac{3}{2} × 0,3= 0,45 (mol)$

⇒ $\begin{cases} m_{AlCl_{3}}=n × M=0,3 × 133,5= 40,05 (g)\\m_{H_{2}}=n × M=0,45 × 2= 0,9 (g) \end{cases}$

Câu 30:

$2Na + 2H_{2}O→2NaOH + H_{2}↑$ (1)

$Fe_{2}O_{3} + 3H_{2}\xrightarrow{t^o}2Fe + 3H_{2}O$ (2)

a) Ta có: $n_{Na}=\dfrac{m}{M}=\dfrac{13,8}{23}= 0,6 (mol)$

Theo PT (1): $n_{NaOH}=n_{Na}= 0,6 (mol)$

Theo PT (1): $n_{H_{2}}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2} × 0,6= 0,3 (mol)$

⇒ $\begin{cases} m_{NaOH}=n × M=0,6 × 40= 24 (g)\\V_{H_{2}}=n × 22,4=0,3 × 22,4= 6,72 (l) \end{cases}$

b) $n_{Fe_{2}O_{3}}=\dfrac{m}{M}=\dfrac{19,2}{160}= 0,12 (mol)$

Lập tỉ lệ: $\dfrac{0,12}{1}>\dfrac{0,3}{3}$

⇒ $Fe_{2}O_{3}$ dư, tính theo mol $H_{2}$

Theo PT (2): $n_{Fe}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3} × 0,3= 0,2 (mol)$

⇒ $m_{Fe}=n × M=0,2 × 56= 11,2 (g)$