Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{MnO_2}=\dfrac{17,4}{87}=0,2\left(mol\right)\\ PTHH:MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ n_{Cl_2\left(TT\right)}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\\ n_{Cl_2\left(LT\right)}=n_{MnO_2}=0,2\left(mol\right)\\ \Rightarrow H=\dfrac{n_{Cl_2\left(TT\right)}}{n_{Cl_2\left(LT\right)}}.100\%=\dfrac{0,16}{0,2}.100=80\%\)
PT: \(2KMnO_4+16HCl\rightarrow2MnCl_2+2KCl+5Cl_2+8H_2O\)
Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
\(m_{HCl}=\dfrac{60.36,5}{100}=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,6}{16}\), ta được KMnO4 dư.
Theo PT: \(n_{Cl_2\left(LT\right)}=\dfrac{5}{16}n_{HCl}=0,1875\left(mol\right)\)
Mà: H% = 80%
\(\Rightarrow n_{Cl_2\left(TT\right)}=0,1875.80\%=0,15\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)
Bạn tham khảo nhé!
MnO\(_2\)+4HCl\(\rightarrow\)MnCl\(_2\)+Cl\(_2\)+2H\(_2O\)
0,45 0,45 (mol)
n\(_{MnO_2}\)=\(\dfrac{39,15}{87}\)=0,45(mol)
2Fe + 3Cl\(_2\)\(\rightarrow\)2FeCl\(_3\)
0,3 0,45 0,3 (mol)
m\(_{FeCl_3}\)=0,3.162,5=48,75(g)
vì hiệu suất phản ứng là 86% nên:
m\(_{FeCl_3}\)=\(\dfrac{86.48,75}{100}\)=41,925(g)
2/
Mg+Cl\(_2\)\(\rightarrow\)MnCl\(_2\)
0,6 0,6
n\(_{Mg}\)=\(\dfrac{14,4}{24}\)=0,6(mol)
2\(KMnO_4+16HCl\rightarrow2MnCl_2+2KCl+5Cl_2\uparrow+8H_2O\)
0,24 0,6
vì hiệu suất phản ứng bằng 80%,nên để điều chế 0,6 mol Cl\(_2\)thì cần số mol \(KMnO_4\) là:
n\(_{KMnO_4}\)=\(\dfrac{0,24.100}{80}\)=0,3(mol)
m\(KMnO_4\)=0,3.158=47,4(g)
bài 17
nKMnO4=23,7\158=0,15 mol
| 16HCl | + | 2KMnO4 | → | 5Cl2 | + | 8H2O | + | 2KCl | + | 2MnCl2 |
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%
\(\Rightarrow m_{KClO3}=\frac{0,078.122,5}{75\%}=12,74\left(g\right)\)
\(n_{KMnO_4}=\frac{15,8}{158}=0,1\left(mol\right)\)
PTHH : \(2KMnO_4+16HCl-->2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(Cl_2+H_2-as->2HCl\) (2)
Có : \(m_{ddHCl}=100\cdot1,05=105\left(g\right)\)
=> \(m_{HCl}=105-97,7=7,3\left(g\right)\)
=> \(n_{HCl}=\frac{7,3}{36,5}=0,2\left(mol\right)\)
BT Clo : \(n_{Cl_2}=\frac{1}{2}n_{HCl}=0,1\left(mol\right)\)
Mà theo lí thuyết : \(n_{Cl_2}=\frac{5}{2}n_{KMnO_4}=0,25\left(mol\right)\)
=> \(H\%=\frac{0,1}{0,25}\cdot100\%=40\%\)
Vì spu nổ thu được hh hai chất khí => \(\hept{\begin{cases}H_2\\HCl\end{cases}}\) (Vì H2 dư)
=> \(n_{hh}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
=> \(n_{H_2\left(spu\right)}=n_{hh}-n_{HCl\left(spu\right)}=0,6-0,2=0,4\left(mol\right)\)
BT Hidro : \(\Sigma_{n_{H2\left(trong.binh\right)}}=n_{H_2\left(spu\right)}+\frac{1}{2}n_{HCl}=0,4+0,1=0,5\left(mol\right)\)
đọc thiếu đề câu a wtf
\(C_{M\left(HCl\right)}=\frac{0,2}{0,1}=2\left(M\right)\)
\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{47,4}{158}=0,3mol\)
\(n_{KMnO_4}=\dfrac{0,3}{80\%}=0,375mol\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
2 16 2 2 5 8 ( mol )
0,375 > 2,5 ( mol )
0,375 0,9375 ( mol )
\(V_{Cl_2}=n_{Cl_2}.22,4=0,9375.22,4=21l\)
\(n_{KMnO_4\left(bd\right)}=\dfrac{47,4}{158}=0,3\left(mol\right)\) => \(n_{KMnO_4\left(pư\right)}=\dfrac{0,3.80}{100}=0,24\left(mol\right)\)
PTHH: 2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,24------------------------------------->0,6
=> \(V=0,6.22,4=13,44\left(l\right)\)
\(m_{MnO_2\left(pư\right)}=\dfrac{43,5.85}{100}=36,975\left(g\right)\)
=> \(n_{MnO_2\left(pư\right)}=\dfrac{36,975}{87}=0,425\left(mol\right)\)
PTHH: MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
0,425-------------------->0,425
=> VCl2 = 0,425.22,4 = 9,52 (l)
MnO2 (0,5 mol) + 4HCl \(\xrightarrow[H=85\%]{t^o}\) MnCl2 + Cl2 (0,425 mol) + 2H2O.
Thể tích khí clo thu được (đktc) là 0,425.22,4=9,52 (lít).