CHƯƠNG II: PHẢN ỨNG HÓA HỌC

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\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3               0,1                   0,3

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ m_{ddAl_2\left(SO_4\right)_3}=5,4+500-0,3.2=504,6g\\ C_{\%A}=C_{\%Al_2\left(SO_4\right)_3}=\dfrac{34,2}{504,6}\cdot100\%=6,78\%\)

`#3107.101107`

Câu `21:`

`a,`

\(\text{4P + 5O}_2\underrightarrow{\text{ }\text{ }\text{ }\text{t}^0\text{ }\text{ }\text{ }}\text{2P}_2\text{O}_5\)

`b,`

\(\text{FeSO}_4+2\text{NaOH}\rightarrow\text{Fe}\left(\text{OH}\right)_2+\text{Na}_2\text{SO}_4\)

Câu `22:`

`a,`

PTHH: \(4\text{Na + O}_2\rightarrow2\text{Na}_2\text{O}\)

`b,`

n của Na trong phản ứng là:

\(\text{n}_{\text{Na}}=\dfrac{\text{m}_{\text{Na}}}{\text{M}_{\text{Na}}}=\dfrac{9,2}{23}=0,4\left(\text{mol}\right)\)

Theo PT: \(\text{n}_{\text{Na}}=2\text{n}_{\text{Na}_2\text{O}}=\dfrac{0,4}{2}=0,2\left(\text{mol}\right)\)

m của Na₂O sau phản ứng là:

\(\text{m}_{\text{Na}_2\text{O}}=\text{n}_{\text{Na}_2\text{O}}\cdot\text{M}_{\text{Na}_2\text{O}}=0,2\cdot\left(23\cdot2+16\right)=0,2\cdot62=12,4\left(\text{g}\right)\)

- Hoặc bạn sử dụng ĐLBT KL:

Theo định luật bảo toàn khối lượng:

\(\text{m}_{\text{Na}}+\text{m}_{\text{O}_2}=\text{m}_{\text{Na}_2\text{O}}\)

`=>`\(\text{m}_{\text{Na}_2\text{O}}=9,2+3,2=12,4\left(\text{g}\right)\)

`c,`

n của O₂ có trong phản ứng là:

\(\text{n}_{\text{O}_2}=\dfrac{\text{m}_{\text{O}_2}}{\text{M}_{\text{O}_2}}=\dfrac{3,2}{16\cdot2}=\dfrac{3,2}{32}=0,1\left(\text{mol}\right)\)

V của O₂ ở đkc là:

\(\text{V}_{\text{O}_2}=\text{n}_{\text{O}_2}\cdot24,79=0,1\cdot24,79=2,479\left(\text{l}\right).\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

1. \(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(n_{Al}=\dfrac{4}{3}n_{O_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)

2. \(n_{KCl\left(25\%\right)}=300.25\%=75\left(g\right)\)

Gọi: m _{dd KCl 10%} = a (g) ⇒ m_{KCl (10%)} = 10%a (g)

\(\Rightarrow\dfrac{75+10\%a}{a+300}=0,15\Rightarrow a=600\left(g\right)\)

`n_{BaCl_2}={20,8}/{208}=0,1(mol)`

`n_{H_2SO_4}={20.19,6\%}/{98}=0,04(mol)`

`H_2SO_4+BaCl_2->BaSO_4+2HCl`

`0,04->0,04->0,04->0,08(mol)`

Do `0,1>0,04->BaCl_2` dư.

`C\%_{BaCl_2\ du}={208(0,1-0,04)}/{20,8+20-0,04.233}.100\%\approx 39,64\%`

`C\%_{HCl}={0,08.36,5}/{20,8+20-0,04.233}.100\%\approx 9,28\%`

a

\(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\downarrow\)

0,05 --> 0,1-----------> 0,05------>0,1

b

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

\(m_{Cu\left(NO_3\right)_2}=0,05.188=9,4\left(g\right)\)

c

\(a=m_{Ag}=108.0,1=10,8\left(g\right)\)

d

\(V_{AgNO_3}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)

\(a.2Al+6HCl->2AlCl_3+3H_2\\ b.m_{AlCl_3}=\dfrac{1}{3}.0,1.0,6.133,5=2,67g\\ c.V_{H_2}=\dfrac{1}{2}.0,06.24,79=0,7437\left(L\right)\\ d.a=\dfrac{1}{3}.0,1.0,6.27=0,54g\)

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

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

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