a) PTHH: \(4Fe+3O_2\rightarrow2Fe_2O_3\)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Xét tỉ lệ ta có:
\(\dfrac{n_{Fe}}{4}>\dfrac{n_{O_2}}{3}\left(\dfrac{0,4}{4}>\dfrac{0,15}{3}\right)\)
\(\Rightarrow Fe\) dư
\(n_{Fe\left(pu\right)}=\dfrac{4\cdot0,15}{3}=0,2\left(mol\right)\)
\(\Rightarrow n_{Fe\left(du\right)}=n_{Fe}-n_{Fe\left(pu\right)}=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe\left(du\right)}=0,2\cdot56=11,2\left(g\right)\)
b) \(m_{Fe\left(du\right)}=11,2\left(g\right)\)
Theo PTHH: \(n_{Fe_2O_3}=\dfrac{0,15\cdot2}{3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,1\cdot160=16\left(g\right)\)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right);n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:4Fe+3O_2\xrightarrow[t^0]{}2Fe_2O_3\)
\(\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Fe.dư\\ n_{Fe\left(pư\right)}=\dfrac{0,15.4}{3}=0,2\left(mol\right)\\ n_{Fe\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ m_{Fe\left(dư\right)}=0,2.56=11,2\left(g\right)\\ PTHH:4Fe+3O_2\xrightarrow[]{t^0}2Fe_2O_3\)
số mol :0,2 0,15 0,1
\(b)m_{Fe_2O_3}=0,1.160=16\left(g\right)\\ m_{chất.rắn}=16+11,2=27,2\left(g\right)\)
