Question

Cho 5,6g \(Fe\) tác dụng với \(O_2\) TẠO \(Fe_2O_3\). Tính \(V_{O_2}=?\left(đktc\right)\);    \(m_{Fe_2O_3}=?\)

Answer 1

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.1.......0.075.....0.05\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(m_{Fe_2O_3}=0.05\cdot160=8\left(g\right)\)

Answer 2

PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)