a/ \(Fe+2HCL\rightarrow FeCl_2+H_2\uparrow\)
\(n_{Fe}=\frac{13,44}{56}=0,24mol\)
100 ml = 0,1 l
\(n_{HCl}=C_M.V_{dd}=2,5.0,1=0,25mol\)
Ta có: \(\frac{0,24}{1}>\frac{0,25}{2}\) vậy Fe dư tính theo số mol HCl
\(n_{FeCl2}=\frac{0,25.1}{2}=0,125mol\)
\(m_{FeCl2}=n.M=0,125.127=15,875g\)
b/ \(n_{H_2}=\frac{0,25.1}{2}=0,125mol\)
\(V_{H_2}=n.22,4=0,125.22,4=2,8l\)
\(\text{Fe + 2HCl }\rightarrow\text{FeCl2 + H2}\)
a)Ta có: nFe=\(\frac{13,44}{56}\)=0,24 mol; nHCl=0,1.2,5=0,25 mol
Nhận thấy 2nFe > nHCl\(\Rightarrow\) Fe dư
\(\rightarrow\) nFe phản ứng=nFeCl2=n\(\frac{HCl}{2}\)=0,125 mol
\(\rightarrow\) mFeCl2=0,125.(56+35,5.2)=15,875 gam
b) Ta có nH2=nFe phản ứng=0,125 mol
\(\rightarrow\) V H2=0,125.22,4=2,8 lít
