Ôn tập học kỳ II

\(N_2O_5+H_2O\rightarrow2HNO_3\left(P.Ứ.hoá.hợp\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\left(P.Ứ.thế\right)\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\left(P.Ứ.hoá.hợp\right)\\ SO_3+H_2O\rightarrow H_2SO_4\left(P.Ứ.hoá.hợp\right)\\ K_2O+H_2O\rightarrow2KOH\left(p.ứ.Hoá.hợp\right)\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\left(P.Ứ.thế\right)\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\left(P.ứ.hoá.hợp\right)\)

- Chất làm quỳ tím chuyển xanh: \(NaOH;Ba\left(OH\right)_2\)

- Chất làm quỳ tím chuyển đỏ: \(HCl;HNO_3;H_2SO_4\)

a, 2Na  +  2H20  ---------> 2NaOH   +   H₂ 

b,H2  +  CuO  ----------->   H₂O + Cu  

c,CaO +  H20 -------------> Ca(0H)2  

d, SO2  +  H20 ------> H₂S0₃ 

a) \(\left\{{}\begin{matrix}n_{Al}=\dfrac{8,64}{27}=0,32\left(mol\right)\\n_{H_2SO_4}=\dfrac{62,72}{98}=0,64\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) Xét tỉ lệ: \(\dfrac{0,32}{2}< \dfrac{0,64}{3}\Rightarrow H_2SO_4\) dư

Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,16\left(mol\right)\\n_{H_2SO_4\left(p\text{ư}\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,48\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{H_2SO_4\left(d\text{ư}\right)}=\left(0,64-0,48\right).96=15,68\left(g\right)\)

c) \(m_{Al_2\left(SO_4\right)_3}=0,16.342=54,72\left(g\right)\)

d) \(V_{H_2}=0,48.22,4=10,752\left(l\right)\)

a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{CuO}=\dfrac{30}{80}=0,375\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,375}{1}>\dfrac{0,3}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Giúp mình với mái thì rồi

\(n_{H_2SO_4}=0,1.3=0,3\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PT: \(n_{Zn}=n_{H_2SO_4}=0,3\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(a/2Zn+O_2\xrightarrow[t^0]{}2ZnO\\ b/P_2O_5+3H_2O\xrightarrow[]{}2H_3PO_4\left(pư.hoá.hợp\right)\\ c/C_2H_4+3O_2\xrightarrow[t^0]{}2CO_2+2H_2O\left(pư.thế\right)\\ d/2K+S\xrightarrow[]{}K_2S\left(pư.hoá.hợp\right)\\ e/2H_2O\xrightarrow[đp]{}2H_2+O_2\left(pư.phân.huỷ\right)\\ f/2KClO_3\xrightarrow[t^0]{}2KCl+3O_2\left(pư.phân.huỷ\right)\\ g/Cu+Cl_2\xrightarrow[]{}CuCl_2\left(pư.hoá.hợp\right)\\ h/2KMnO_4\xrightarrow[t^0]{}K_2MnO_4+MnO_2+O_2\left(pư.phân.huỷ\right)\\ i/Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\left(pư.thế\right)\\ j/2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\left(pư.thế\right)\\ k/Sai.đề\\ l/CaO+H_2O\xrightarrow[]{}Ca\left(OH\right)_2\left(pư.hoá.hợp\right)\)