Đề kiểm tra 1 tiết chương I: Oxit-Axit -Đề 1

1A

2C

3C

4A

5D

6/

a)\(2Cu + O_2 \rightarrow^{t^o} 2CuO\)

b)\(CuCl_2 + 2NaOH \rightarrow Cu(OH)_2 + 2NaCl\)

c)\(Fe_2(SO_4)_3 + 3BaCl_2 \rightarrow 3BaSO_4 + 2FeCl_3\)

d)\(2Fe(OH)_3 \rightarrow^{t^o} Fe_2O_3 + 3H_2O\)

e)\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)

\(a,n_{MgCl_2}=\dfrac{38}{24+35,5\cdot2}=0,4\left(mol\right)\\ PTHH:MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\\ TL:......1........2........1.......2\\ BR:.....0,4.......0,8........0,4........0,8\left(mol\right)\)

\(b,m_{Mg\left(OH\right)_2}=0,4\cdot\left(24+17\cdot2\right)=23,2\left(g\right)\\ c,PTHH:Mg\left(OH\right)_2\rightarrow^{t^0}MgO+H_2O\)

Theo PTHH ta có \(n_{Mg\left(OH\right)_2}=n_{MgO}=0,4\left(mol\right)\)

\(\Rightarrow m_{đề}=m_{MgO}=40\cdot0,4=16\left(g\right)\)

a)

$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$
$Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$

b)

$n_{Mg(OH)_2} = n_{MgCl_2} = \dfrac{38}{95}  = 0,4(mol)$
$m_{Mg(OH)_2} = 0,4.58 = 23,2(gam)$

c)

$n_{MgO} = n_{MgCl_2} = 0,4(mol)$
$m_{MgO} = 0,4.40 = 16(gam)$

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

a)\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

    0,3      0,3            0,3          0,3

b) \(m_{Fe}=0,3\cdot56=16,8\left(g\right)\)

c) \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,3}{0,15}=2M\)

    \(m_{FeSO_4}=0,3\cdot\left(56+32+4\cdot16\right)=45,6\left(g\right)\)

Do Cu không pư HCl

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

\(n_{H_2}= \dfrac{1,12}{22,4}= 0,05 mol\)

Theo PTHH:

\(n_{Zn}= n_{H_2}= 0,05 mol\)

\(\Rightarrow m_{Zn}= 0,05 . 65=3,25g\)

\(\rightarrow m_{Cu}= 6,45 - 3,25 =3,2 g\)

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14c

15b

16d

17b

18d

19b

20d

21c

22b

23a