\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O \\ Mol:0,1\rightarrow0,1\\ m_{CuO}=0,1.80=8\left(g\right)\\ \Rightarrow B\)

d

D

a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)

b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)

Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

\(m_{CuO}=0,5.80=40\left(g\right)\)

a) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,05\cdot98=4,9\left(g\right)\)

b) PTHH: \(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,05\left(mol\right)\\n_{H_2SO_4}=\dfrac{250\cdot9,8\%}{98}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{CuSO_4}=0,05\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,05\cdot160}{4,9+250}\cdot100\%\approx3,14\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2\cdot98}{4,9+250}\cdot100\%\approx7,7\%\end{matrix}\right.\) 

n HCl = 360 x 18,25/(100x36,5) = 1,8 mol

H 2  + CuO  → t ° Cu +  H 2 O

n CuO  = x

Theo đề bài

m CuO (dư) +  m Cu =  m CuO  (dư) +  m Cu   p / u  - 3,2

m Cu  =  m Cu   p / u  - 3,2 => 64x = 80x - 3,2

=> x= 0,2 mol →  m H 2  = 0,4g

Fe + 2HCl → FeCl 2  +  H 2

Số mol HCl tác dụng với  Fe 3 O 4 ,  Fe 2 O 3 , FeO là 1,8 - 0,4 = 1,4 mol

Phương trình hóa học của phản ứng:

Fe 3 O 4 + 8HCl → 2 FeCl 3 + FeCl 2  + 4 H 2 O (1)

Fe 2 O 3  + 6HCl → 2 FeCl 3  + 3 H 2 O  (2)

FeO + 2HCl →  FeCl 2  +  H 2 O  (3)

Qua các phản ứng (1), (2), (3) ta nhận thấy  n H 2 O  = 1/2 n HCl  = 1,4:2 = 0,7 mol

Áp dụng định luật bảo toàn khối lượng, ta có:

m hỗn   hợp  +  m HCl  =  m muối  +  m H 2 O  +  m H 2

57,6 + 1,8 x 36,5 =  m muối  + 0,7 x 18 +0,4

m muối  = 57,6 + 65,7 - 12,6 - 0,4 = 110,3 (gam)

Ta có : n_Cu(OH)2=19,6/98=0,2 mol

PTHH:

Cu(OH)2 →t^o→ CuO + H2O

⇒n_Cu(OH)2=n_CuO=0,2 mol

m_CuO=0,2.80=16g

Đáp án:C

\(Cu\left(OH\right)_2\rightarrow CuO+H_2O\)

\(\Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

a, \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

PT: \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuO}=0,05\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,05.98=4,9\left(g\right)\)

b, \(m_{H_2SO_4}=250.9,8\%=24,5\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

PT: \(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,25}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{CuSO_4}=n_{H_2SO_4\left(pư\right)}=n_{Cu\left(OH\right)_2}=0,05\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,25-0,05=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,05.160}{4,9+250}.100\%\approx3,14\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{4,9+250}.100\%\approx7,69\%\end{matrix}\right.\)

Ta có: \(n_{KClO_3}=\dfrac{49}{122,5}=0,4\left(mol\right)\)

PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

______0,4______0,4_____0,6 (mol)

\(\Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)

\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)

a/ Ta có: \(n_{KClO_3}=\dfrac{12.25}{122.5}=0.1\left(mol\right)\)

PTHH:

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)

2                              3

0.1                           x

\(=>x=\dfrac{0.1\cdot3}{2}=0.15=n_{O_2}\)

\(=>V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)

PTHH: \(KClO_3\underrightarrow{t^o}KCl+\dfrac{3}{2}O_2\)

a) Ta có: \(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

\(\Rightarrow n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)

b) PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,15}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết

\(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\)

\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}\cdot0,1=0,15\left(mol\right)\\ V_{O_2\left(ĐKTC\right)}=n_{O_2}\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\\ n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \dfrac{n_{Fe}}{n_{O_2}}=2>\dfrac{3}{2}\Rightarrow Fe\text{ dư, bài toán tính theo lượng }O_2\\ m_{Fe_3O_4}=n_{Fe_3O_4}\cdot M_{Fe_3O_4}=0,075\cdot232=17,4\left(g\right)\)