1/

$1.FeO+H_2\xrightarrow{t^o} Fe+H_2O$

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

$3.Fe_3O_4+4H_2\xrightarrow{t^o} 3Fe+4H_2O$

$4.2H_2+O_2\xrightarrow{t^o} 2H_2O$

$5.2Ba+O_2\xrightarrow{t^o} 2BaO$

$6.4Na+O_2\xrightarrow{t^o} 2Na_2O$

$7.Ba+2H_2O\xrightarrow{} Ba(OH)_2+H_2↑$

$8.SO_2+H_2O\xrightarrow{} H_2SO_3$

2/

$2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$

$CaO+H_2O\xrightarrow{} Ca(OH)_2$

$BaO+H_2O\xrightarrow{} Ba(OH)_2$

$P_2O_5+3H_2O\xrightarrow{} 2H_3PO_4$

$Ca+H_2O\xrightarrow{} Ca(OH)_2+H_2↑$

$SO_3+H_2O\xrightarrow{} H_2SO_4$

3/

Oxit axit: $SO_3$

Muối: $Fe_2(SO_4)_3,MgSO_4$

Bazơ: $KOH$

Axit: $HCl,HNO_3$

(Lưu ý: Bạn sửa "axit liric" thành "axit nitric" nha)

4/

$a,C\%_{KCl}=\dfrac{5}{200}.100\%=2,5\%$

$b,CM_{NaCl}=\dfrac{0,5}{2}=0,25M.$

$c,m_{ct}(NaCl)=300.10\%=30g.$

$⇒m_{H_2O}=300-30=170g.$

5/

$PTPƯ:Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$

$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$

$m_{HCl}=300.4,9\%=14,7g.$

$⇒n_{HCl}=\dfrac{14,7}{36,5}≈0,4mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{1}<\dfrac{0,4}{1}$

$⇒n_{H_2SO_4}$ $dư,$ $\text{tính theo}$ $n_{Fe}$

$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,1mol.$

$⇒V=V_{H_2}=0,1.22,4=2,24l.$

chúc bạn học tốt!

Câu 1 :

1, FeO+H₂ ⟶ Fe+H₂O

2, Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

3, Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

4, 2H₂ + O₂ → 2H₂O

5,  2Ba + O₂ → 2BaO

6,  4Na + O₂ → 2Na₂O

7, Ba + 2H₂O → Ba(OH)₂ + 2H

8, SO₂ + H₂O → H₂SO₃