Ôn tập học kỳ II

a)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

b)

n CuO = 2,8/80 = 0,035(mol)

n H2SO4 = 140.20%/98  = 2/7 (mol)

Ta thấy : 

n CuO / 1 < n H2SO4 /1 nên H2SO4 dư

n H2SO4 pư = n CuO = 0,035(mol)

=> m H2SO4 dư = 140.20% - 0,035.98 = 24,57(gam)

c)

m dd = m CuO + m dd H2SO4 = 2,8 + 140 = 142,8(gam)

C% CuSO4 = 0,035.160/142,8   .100% = 3,92%

C% H2SO4 = 24,57/142,8   .100% = 17,2%

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\left(1\right)\)

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

Từ phương trình phản ứng ta có :

\(n_{H_2SO_4}=n_{Na_2O}=0,1\left(mol\right)\)

\(C_M\left(H_2SO_4\right)=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5M\)

\(\left(1\right)\Rightarrow n_{Na_2SO_4}=0,1\left(mol\right)\)

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

1)

PT: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

Ta có: \(n_{H_2SO_4}=\dfrac{a.C\%}{98}\left(mol\right)\)

m_{H2O (trong acid)} = a - a.C% \(\Rightarrow n_{H_2O}=\dfrac{a-a.C\%}{18}\left(mol\right)\)

\(n_{H_2}=\dfrac{0,05a}{2}=0,025\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}+\dfrac{1}{2}n_{H_2O}=n_{H_2}\)

\(\Rightarrow\dfrac{a.C\%}{98}+\dfrac{1}{2}.\dfrac{a-a.C\%}{18}=0,025a\Rightarrow C=15,8\%\)

2)

PT: \(MO+H_2SO_4\rightarrow MSO_4+H_2O\)

Ta có: \(n_{MO}=\dfrac{b}{M_M+16}\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{MSO_4}=n_{MO}=\dfrac{b}{M_M+16}\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{\dfrac{b}{M_M+16}.98}{15,8\%}=\dfrac{620,25b}{M_M+16}\left(g\right)\)

⇒ m _{dd sau pư} = b + 620,25b/(M_M + 16) (g)

\(\Rightarrow C\%_{MSO_4}=\dfrac{\dfrac{b}{M_M+16}.\left(M_M+96\right)}{b+\dfrac{620,25b}{M_M+16}}.100\%=18,21\%\)

⇒ M_M = 24 (g/mol)

→ M là Mg.

Pt: Fe + 2HCl => FeCl₂ + H₂

nFe = \(\dfrac{2,8}{56}=0,05mol\)

nHCl = \(\dfrac{14,6}{36,5}=0,4mol\)

nFe : nHCl = \(\dfrac{0,05}{1}:\dfrac{0,4}{2}=0,05:0,2=1:4\)

=> HCl dư

a) nFeCl₂ = nFe = 0,05 mol => mFeCl₂ = 6,35g

b) nH₂ = nFe = 0,05 mol => VH₂ = 0,05.22,4 = 1,12 lít

Bài 1:

\(1.\left(x+1\right)^2=x^2+2\cdot x\cdot1+1^2=x^2+2x+1\\ 2.\left(4+x\right)^2=4^2+2\cdot4\cdot x+x^2=16+8x+x^2\\ 3.\left(6-x\right)^2=6^2-2\cdot6\cdot x+x^2=36-12x+x^2\\ 4.\left(x-5\right)^2=x^2-2\cdot x\cdot5+5^2=x^2-10x+25\\ 5.\left(5x+1\right)^2=\left(5x\right)^2+2\cdot5x\cdot1+1^2=25x^2+10x+1\\ 6.\left(2x+3\right)^2=\left(2x\right)^2+2\cdot2x\cdot3+3^2=4x^2+12x+9\\ 7.\left(2x-1\right)^2=\left(2x\right)^2-2\cdot2x\cdot1+1^2=4x^2-4x+1\\ 8.\left(3x-2\right)^2=\left(3x\right)^2-2\cdot3x\cdot2+2^2=9x^2-12x+4\\ 9.\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\\ 10.\cdot\left(x+5y\right)^2=x^2+2\cdot x\cdot5y+\left(5y\right)^2=x^2+10xy+25y^2\\ 11.\left(x-2y\right)^2=x^2-2\cdot x\cdot2y+\left(2y\right)^2=x^2-4xy+4y^2\\ 12.\left(2x-y\right)^2=\left(2x\right)^2-2\cdot2x\cdot y+y^2=4x^2-4xy+y^2\)

Bài 4:

1: \(x\left(1-x\right)+\left(x-1\right)^2\)

\(=x-x^2+x^2-2x+1\)

=-x+1

2: \(\left(x-3\right)^2-x^2+10x-7\)

\(=x^2-6x+9-x^2+10x-7\)

=4x+2

3: \(\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)\)

\(=x^2+4x+4-\left(x^2-2x-3\right)\)

\(=x^2+4x+4-x^2+2x+3\)

=6x+7

4: \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2\)

\(=x^2-2x+4x-8-\left(x^2-6x+9\right)\)

\(=x^2+2x-8-x^2+6x-9\)

=8x-17

Bài 2:

1: \(x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)

2: \(1-4x^2=1-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)

3: \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

4: \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5: \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

6: \(9x^2-36=9\left(x^2-4\right)=9\left(x-2\right)\left(x+2\right)\)

\(3.\\ n_{ZnO}=\dfrac{8,1}{81}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1       0,2            0,1           0,1

\(m_{ZnCl_2}=0,1.136=13,6g\\ C_{\%HCl}=\dfrac{0,2.36,5}{120}\cdot100\%=6,08\%\\ 4.\\ n_K=\dfrac{7,8}{39}=0,2mol\\ 2K+2H_2O\rightarrow2KOH+H_2\)

0,2       0,2             0,2           0,1

\(KOH+HCl\rightarrow KCl+H_2O\)

0,2            0,2            0,2           0,2  

\(m_{KCl}=0,2.74,5=14,9\\ C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100\%=3,65\%\)

\(n_{Al}=\dfrac{10,8}{27}=0,4mol\\ 2Al+3S\xrightarrow[t^0]{}Al_2S_3\\ n_{Al_2S_3\left(lt\right)}=\dfrac{1}{2}n_{Al}=0,2mol\\ m_{Al_2S_3\left(lt\right)}=0,2.150=30g\\ H=\dfrac{25,5}{30}\cdot100\%=85\%\)
 

\(a.500ml=0,5l\\ C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,5}=0,2M\\ b.m_{ddNa_2CO_3}=500.1=500g\\ m_{Na_2CO_3}=0,1.106=10,6g\\ C_{\%Na_2CO_3}=\dfrac{10,6}{500}\cdot100\%=2,12\%\)

NaF: natri florua

CuCl₂: đồng(ll) clorua

CaSO₃: canxi sunfit

Ba(HCO₃)₂: bari hidrocacbonat

KHSO₄: kali hidrosunfat

\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)

                   1          2               1            1

                  0,2       0,4            0,2         0,2 (mol)

\(a)V_{H_2}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

\(b)m_{ZnCl_2}=n\cdot M=0,2\cdot\left(65+35,5\cdot2\right)=27,2\left(g\right)\)

\(c)400ml=0,4l\\ C_{M_{HCl}}=\dfrac{n}{V_{dd}}=\dfrac{0,4}{0,4}=1M.\)

a) PTHH:

Zn + 2HCl -> ZnCl2 + H2

0,2     0,4         0,2       0,2

nZn = 13 : 65 = 0,2 (mol)

VH2 = 0,2 . 22,4 = 4,48 (l)

b) mZnCl2 = 0,2 . 201 = 40,2 (g)

c) CM HCl = 0,4 : 0,4 = 1 (M)