bài 17

nKMnO4=23,7\158=0,15 mol

2KMnO4 + 16HCl = 2KCl + 2MnCl2 + 5Cl2 + 8H2O

0,6 ----------------------- 0,1875 mol

nHCl=21,9\36,5=0,6 mol

=>VCl2=0,1875.22,4=4,2 l

\(\left(1\right)\) \(n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\)

\(PTHH:2KMnO_4+16HCl_{\left(\text{đ}\right)}\underrightarrow{t^o}2KCl+2MnCl_2+5Cl_2+8H_2O\)

(mol)________________0,6__________________________3/16___________

\(V_{Cl_2}=22,4.\frac{3}{16}=4,2\left(l\right)\)

\(\left(2\right)\) \(n_{KMnO_4}=\frac{23,7}{158}=0,15\left(mol\right)\)

\(PTHH:2KMnO_4+16HCl_{\left(\text{đ}\right)}\underrightarrow{t^o}2KCl+2MnCl_2+5Cl_2+8H_2O\)

(mol)_______0,15________________________________0,375____________

\(V_{Cl_2}=22,4.0,375=8,4\left(l\right)\)

Chọn C

Chọn C.

Dung dịch Y chứa H⁺ với n H + = 2 n H 2 = 0 , 12   mol ⇒ H + ban đầu:  0 , 12 + 2 n H 2 ⇒ n O 2 = 0 , 03 + 0 , 5 n H 2

Khối lượng dung dịch giảm:

m Cu + m khí = 300 - 283 , 32 = 16 , 68 ⇒ 64 n Cu + 71 n Cl 2 + 18 n H 2 = 15 , 72  (1)

→ BT :   e 2 n Cu = 0 , 12 + 2 n Cl 2 (2) và

  160 n Cu + 117 n Cl 2 = 31 , 76 (3)

Từ (1), (2), (3) suy ra: 

n Cu =   0 , 14   mol ;   n Cl 2 =   0 , 08   mol   ; n H 2 = 0 , 06   mol ⇒ n O 2 = 0 , 06   mol

Vậy V = 3,136 lít.

Bài 43:\(PTHH:2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\left(1\right)\)

\(Cl_2+2NaOH\rightarrow NaClO+NaC;+H_2O\left(2\right)\)

Ta có :

\(m_{NaCl}+m_{NaClO}=33,25\left(mol\right)\)

Theo PTHH \(n_{NaCl}=n_{NaClO}=a\left(mol\right)\)

\(\Rightarrow58,5a+74,5=33,25\)

\(\Rightarrow a=0,25\left(mol\right)\)

Theo PTHH (2) \(n_{Cl2}=n_{NaCl}=0,25\left(mol\right)\)

Theo PTHH (1)\(\Rightarrow n_{KMnO4}=\frac{2}{5}n_{Cl2}=\frac{2}{5}.0,25=0,1\left(mol\right)\)

Mà H điều chế = 80% \(\Rightarrow n_{KMnO4\left(bđ\right)}=\frac{0,1}{80\%}=0,125\left(mol\right)\)

\(\Rightarrow m_{KMnO4\left(bđ\right)}=0,125.155=19,375\left(g\right)\)

Bài 44:

\(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)

\(Cl_2+2KOH\rightarrow KCl+KClO+H_2O\)

Ta có:

\(n_{KCl}=\frac{17,433}{74,5}=0,234\left(mol\right)\)

\(\Rightarrow n_{Cl2}=n_{KCl}=0,234\left(mol\right)\)

\(\Rightarrow n_{KClO3}=\frac{0,234}{3}=0,078\left(mol\right)\)

Mà H =75%

\(\Rightarrow m_{KClO3}=\frac{0,078.122,5}{75\%}=12,74\left(g\right)\)

Đáp án C

Cr₂O₃ + 2Al → t 0  Al₂O₃ + 2Cr

Ÿ Phần 2: + NaOH đặc nóng →  0,075 mol H₂

Ÿ Phần 1: + HCl loãng, nóng → 0,15 mol H₂

_{Có  0 , 1 1 < 0 , 25 2} => tính hiệu suất theo lượng Cr₂O₃ phản ứng.

Chọn đáp án B

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{HCl}=0,18\cdot1=0,18\left(mol\right)\)

\(\Rightarrow n_{H_2\left(LT\right)}=0,09\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{\dfrac{1,512}{22,4}}{0,09}\cdot100\%=75\%\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{HCl}=0,18.1=0,18\left(mol\right)\)

Theo PT: \(n_{H_2\left(LT\right)}=\dfrac{1}{2}n_{HCl}=0,09\left(mol\right)\)

\(\Rightarrow V_{H_2\left(LT\right)}=0,09.22,4=2,016\left(l\right)\)

Mà: V_{H2 (TT)} = 1,512 (l)

\(\Rightarrow H\%=\dfrac{1,512}{2,016}.100\%=75\%\)

Bạn tham khảo nhé!