\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ \text{Đ}LBTKL:m_{NaOH}+m_{H_2SO_4}=m_{Na_2SO_4}+m_{H_2O}\\ \Leftrightarrow80+98=m_{Na_2SO_4}+36\\ \Leftrightarrow m_{Na_2SO_4}=142\left(kg\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

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

b 

\(a=m_{Cu}=0,1.64=6,4\left(g\right)\)

c

\(V_{CuSO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)

a) Dấu hiệu : Xuất hiện chất rắn màu trắng

b) $Na_2CO_3 + Ca(OH)_2 \to CaCO_3 + 2NaOH$

c) $m_{Na_2CO_3} + m_{Ca(OH)_2} = m_{CaCO_3} + m_{NaOH}$

1) \(2KClO_3\rightarrow2KCl+3O_2\)

2) 

a) \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

b) \(2Na+2H_2O\rightarrow2NaOH+H_2\)

c) \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

1) 2KClO₃ --t^o--> 2KCl + 3O₂

2a) CuO + H₂ --t^o--> Cu + H₂O

b) 2Na + 2H₂O ----> 2NaOH + H₂

c) Zn + H₂SO₄ ----> ZnSO₄ + H₂

a. PTK_CO = 12 + 16 = 28(đvC)

b. \(PTK_{H_2SO_4}=1.2+32+16.4=98\left(đvC\right)\)

c. \(PTK_{Cu\left(OH\right)_2}=64+\left(16+1\right).2=98\left(đvC\right)\)

d. \(PTK_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)

e. \(PTK_{C_6H_{12}O_6}=12.6+1.12+16.6=180\left(đvC\right)\)

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,1.2=0,2mol\\ C_{M_X}=C_{M_{NaOH}}=\dfrac{0,2}{2}=0,1M\)

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{2}=0,1\left(m\right)\)

- Phi kim: S, P và H

- Kim loại: Cu, Na và Fe