1.\(n_{HCl}=2,5.0,1=0,25mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,25 0,125 ( mol )
\(V_{H_2}=0,125.22,4=2,8l\)
2.\(n_{HCl}=\dfrac{7,3}{36,5}=0,2mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 0,1 ( mol )
\(V_{H_2}=0,1.22,4=2,24l\)
\(n_{HCl}=0,1.2,5=0,25\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,25 0,125
\(V_{H_2}=0,125.22,4=2,8L\)
2 .
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,2 0,1
\(V_{H_2}=0,1.22,4=2,24L\)