Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
______0,5____1______0,5____0,5 (mol)
b, Ta có: \(m_{HCl}=1.36,5=36,5\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{36,5}{20\%}=182,5\left(g\right)\)
⇒ m dd sau pư = 28 + 182,5 - 0,5.2 = 209,5 (g)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{0,5.127}{209,5}.100\%\approx30,31\%\)
Bạn tham khảo nhé!
\(a)Fe+2HCl\rightarrow FeCl2+H2\)
\(b)n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(n_{HCl}=2n_{Fe}=1\left(mol\right)\)
\(\Rightarrow m_{HCl}=36,5\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{36,5.100}{20}=182,5\left(g\right)\)
\(n_{H2}=n_{Fe}=0,5\left(mol\right)\Rightarrow m_{H2}=1\left(g\right)\)
\(m_{dd}saupư=182.5+28-1=209,5\left(g\right)\)
\(n_{FeCl2}=n_{Fe}=0,5\left(mol\right)\)
\(C\%_{FeCl2}=\dfrac{0,5.127}{209,5}.100\%=30,3\%\)
Chúc bạn học tốt^^
