Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{AgCl}=\dfrac{7,175}{143,5}=0,05\left(mol\right)\)
PTHH: HCl + AgNO3 ---> AgCl↓ + HNO3
0,05<---------------0,05
\(\rightarrow m_{HCl}=0,05.36,5=1,825\left(g\right)\\
\rightarrow C\%_{ddA}=\dfrac{1,825}{50}.100\%=3,65\%\)
\(n_{Cl_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Đặt H = x%
PTHH: Cl2 + H2 --as--> 2HCl
LTL: 6,72 < 10 => H2 dư
=> nHCl = 0,3x (mol)
\(\rightarrow C\%_{HCl}=\dfrac{0,3x.36,5}{0,3x.36,5+385,4}.100\%=3,65\%\\ \Leftrightarrow20,23\%\)
\(n_{H_2}=\dfrac{1}{22,4}=\dfrac{5}{112}\left(mol\right)\)
\(n_{Cl_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: H2 + Cl2 --as--> 2HCl
Xét tỉ lệ: \(\dfrac{\dfrac{5}{112}}{1}>\dfrac{0,03}{1}\) => Hiệu suất tính theo Cl2
\(n_{AgCl}=\dfrac{7,175}{143,5}=0,05\left(mol\right)\)
Bảo toàn Cl: \(n_{Cl_2\left(pư\right)}=0,025\left(mol\right)\)
\(H\%=\dfrac{0,025}{0,03}.100\%=83,33\%\)
\(n_{AgCl}=\dfrac{43.05}{143.5}=0.3\left(mol\right)\) \(\Rightarrow n_{HCl}=0.3\left(mol\right)\)
\(n_{HCl}=\dfrac{6.72}{22.4}=0.3\left(mol\right),n_{Cl_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(H_2+Cl_2\underrightarrow{^{^{t^0}}}2HCl\)
\(0.15....0.15.......0.3\)
\(H\%=\dfrac{0.15}{0.2}\cdot100\%=75\%\)
\(H_2+Cl_2\rightarrow2HCl\)
\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)
\(n_{H2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)
\(n_{CL2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)
\(\rightarrow\) Tính theo Cl2
\(n_{AgCl}=\frac{1,435}{143,5}=0,01\left(mol\right)\)
\(\rightarrow n_{HCl_{bđ}}=0,02\left(mol\right)\)
\(\rightarrow n_{HCl_{lt}}=0,025.2=0,05\left(mol\right)\)
\(\rightarrow H=\frac{0,02}{0,05}.100=40\%\)
Câu 1:
\(H_2+Cl_2\rightarrow2HCl\left(1\right)\)
___0,03______0,06
\(HCl+AgNO_3\rightarrow AgCl+HNO_3\)
\(n_{AgCl}=\frac{0,17}{108+35,5}\)
\(n_{HCl\left(1\right)}=\frac{0,17}{108+35,5}.\frac{20}{5}=\frac{34}{7175}\)
\(n_{H2}=\frac{1}{22,4}\)
\(n_{Cl2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)
\(\Rightarrow H=\frac{\frac{34}{7145}}{0,06}.100\%=7,9\%\)
Câu 3:
\(n_{NaCl}=\frac{12,87}{23+35,5}=0,22\left(mol\right)\)
\(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
0,22____________0,11___________0,22
\(n_{NaCl}=n_{HCl}=0,22\left(mol\right)\)
\(\Rightarrow V_{HCl}=0,22.22,4.90\%=4,43521\)
\(m_{Na2SO4}=0,11.142.90\%=14,058\left(g\right)\)