a. \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)
PTHH : 2KClO3 -to> 2KCl + 3O2
0,2 0,3
\(V_{O_2}=0,3.22,4=6,72\left(l\right)\)
b.\(n_{O_2}=0,3\left(mol\right)\left[cmt\right]\)
\(m_{O_2}=0,3.32=9,6\left(g\right)\)
PTHH : 3Fe + 2O2 -to> Fe3O4
Theo ĐLBTKL
\(m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow11,2+9,6=20,8\left(g\right)\)
a: \(2KClO_3\rightarrow2KCl+3O_2\)
\(n_{KClO_3}=\dfrac{24.5}{122.5}=0.2\left(mol\right)\)
\(\Leftrightarrow n_{O_2}=\dfrac{3}{2}\cdot0.2=0.3\left(mol\right)\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(lít\right)\)
b: \(4Fe+3O_2\rightarrow2Fe_2O_3\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(\dfrac{n_{Fe}}{4}=\dfrac{0.2}{4}=0.05< \dfrac{n_{O_2}}{3}\) nên O2 dư
=>Tính theo Fe
\(n_{Fe_2O_3}=0.1\left(mol\right)\)
\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)
