PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
Bạn tham khảo nhé!
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) HCl phản ứng hết, Fe còn dư
\(\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\) \(\Rightarrow m_{Fe\left(dư\right)}=0,05\cdot56=2,8\left(g\right)\)
b) Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol\)
\(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)