Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tham khảo:
nNaCl=117 : 58,5 = 2(mol)
2NaCl +2H2O→2NaOH+H2↑+Cl2↑
=> nCl2 = 2x1:2=1(mol)
=> mCl2= 35,5 x 2 = 71 (g)
mCl2 thực tế = 71 x 80% = 56,8 (g)
nNaCl = 117/58,5 = 2 (mol)
PTHH: 2NaCl + 2H2O -> (đpcmn) 2NaOH + Cl2 + H2
nCl2 (LT) = 2/2 = 1 (mol)
nCl2 (TT) = 1 . 80% = 0,8 (mol)
mCl2 (TT) = 0,8 . 71 = 56,8 (g)
2NaCl +2H2O→2NaOH+H2↑+Cl2↑
H%=80=\(\frac{2n_{Cl_2}}{n_{Nacl}}\) →mCLo
20 tấn = 20 000 kg
\(m_{NaCl} = 20 000.90\% = 18000(kg)\\ n_{NaCl} = \dfrac{18000}{58,5}= \dfrac{4000}{13}(kmol)\\ n_{NaCl\ pư} = \dfrac{4000}{13}.65\% = 200(kmol)\\ 2NaCl + 2H_2O \xrightarrow{đpdd} 2NaOH + Cl_2 + H_2\\ n_{Cl_2} = \dfrac{1}{2}n_{NaCl} = 100(kmol)\\ V_{Cl_2} = 100.22,4 = 2240(m^3)\)
\(a,2KMnO_4+16HCl_{đặc}\rightarrow\left(t^o\right)2KCl+2MnCl_2+5Cl_2+8H_2O\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Ta.có:n_{FeCl_3}=\dfrac{39}{162,5}=0,24\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,24\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,24=0,36\left(mol\right)\\ n_{K_2MnO_4}=\dfrac{2}{5}.0,36=0,144\left(mol\right)\\ n_{HCl}=\dfrac{16}{5}.0.36=1,152\left(mol\right)\\ \Rightarrow a=m_{KMnO_4}=0,144.158=22,752\left(g\right)\\ b=C_{MddHCl}=\dfrac{1,152}{0,1}=11,52\left(M\right)\\ x=m_{Fe}=0,24.56=13,44\left(g\right)\\ V=V_{Cl_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right) \)
\(b,n_{KCl}=n_{MnCl_2}=\dfrac{2}{5}.0,36=0,144\left(mol\right)\\ KCl+AgNO_3\rightarrow AgCl\downarrow\left(trắng\right)+KNO_3\\ MnCl_2+2AgNO_3\rightarrow2AgCl\downarrow\left(trắng\right)+Mn\left(NO_3\right)_2\\ n_{AgNO_3}=n_{AgCl}=n_{KCl}+2.n_{MnCl_2}=0,144+2.0,144=0,432\left(mol\right)\\ \Rightarrow m_{AgCl\downarrow\left(trắng\right)}=143,5.0,432=61,992\left(g\right)\\ m_{AgNO_3}=0,432.170=73,44\left(g\right)\\ \Rightarrow m_{ddAgNO_3}=\dfrac{73,44.100}{5}=1468,8\left(g\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)
a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)
c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
Bạn tham khảo nhé!
Đáp án C
MnO2 + 4HCl →MnCl2 + 2H2O + Cl2
0,1 →0,1 (mol)
Do H% = 85% => 
= 0,085 (mol)
V = 0,085.22,4 = 1,904 (lít)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Pt : \(2NaCl+2H_2O\xrightarrow[có.màng.ngăn]{điện.phân}2NaOH+H_2+Cl_2|\)
2 2 2 1 1
0,6 0,3
\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3|\)
2 3 2
0,2 0,3
\(n_{NaCl}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)
⇒ \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
Chúc bạn học tốt
nNaCl = 1.0,5=0,5(mol)
Ta có PT:
2NaCl + 2H2O ---> 2NaOH + Cl2 + H2
....0,5........0,5...............0,5.......0,25..0,25
Vì hiệu suất pư xảy ra là 80%
=> n\(Cl_2\) thu được =0,25.80%=0,2(mol)
Ta có PT
2Fe + 3Cl2 ---> 2FeCl3
nFe=\(\frac{11,2}{56}\)=0,2(mol)
Ta có: \(\frac{0,2}{2}\)>\(\frac{0,2}{3}\)=> Fe dư
=> n\(FeCl_3\)=\(\frac{2}{3}\)n\(Cl_2\)=\(\frac{2}{3}\).0.2=0,1333(mol)
m\(FeCl_3\)=0,1333.162,5=21,66(g)