a)\(3Fe+2O2--.Fe3O4\)
\(n_{Fe}=\frac{84}{56}=1,5\left(mol\right)\)
\(n_{O2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
Lập tỉ lệ
\(n_{Fe}\left(\frac{1,5}{3}\right)< n_{O2}\left(\frac{1,5}{2}\right)=>O2dư\)
\(n_{O2}=\frac{2}{3}n_{Fe}=0,1\left(mol\right)\)
\(n_{O2}dư=0,15-0,1=0,05\left(mol\right)\)
\(m_{O2}dư=0,05.32=1,6\left(g\right)\)
b)\(n_{Fe3O4}=\frac{1}{3}n_{Fe}=0,05\left(mol\right)\)
\(mFe3O4=0,05.232=11,6\left(g\right)\)
a,
\(n_{Fe}=\frac{84}{56}=1,5\left(mol\right)\)
\(n_{O2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(PTHH:3Fe+2O_2\rightarrow Fe_3O_4\)
Tỉ lệ : \(\frac{1,5}{3}< \frac{1,5}{2}\)
Nên O2 dư , Fe hết
Ta có :
\(n_{O2}=0,1\left(mol\right)\Rightarrow n_{O2\left(Dư\right)}=0,15-0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{O2\left(Dư\right)}=0,05.32=1,6\left(g\right)\)
b,\(n_{Fe3O4}=0,05\left(mol\right)\Rightarrow m_{Fe3O4}=0,05.232=11,6\left(g\right)\)
