Bài 15:
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a+b) Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,3\left(mol\right)\\n_{Fe_3O_4}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\m_{Fe_3O_4}=0,1\cdot232=23,2\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,4\cdot158=63,2\left(g\right)\)
Bài 6:
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{1}{2}\) \(\Rightarrow\) Fe p/ứ hết, HCl còn dư
\(\Rightarrow n_{HCl\left(dư\right)}=1-0,2=0,8\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,8\cdot36,5=29,2\left(g\right)\)
b+c) Theo PTHH: \(n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
