\(n_{Fe}=\dfrac{m}{M}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
1 : 2 : 1 (mol)
0,05 : 0,4 (mol)
-Chuyển thành tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,4}{2}\Rightarrow\) Fe phản ứng hết còn HCl dư.
-Theo PTHH: \(n_{H_2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
b) \(n_{Fe\left(cần\right)}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(\Rightarrow n_{Fe\left(thêm\right)}=n_{Fe\left(cần\right)}-n_{Fe\left(tt\right)}=0,2-0,05=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe\left(thêm\right)}=n.M=0,15.56=8,4\left(g\right)\)
