\(n_{Mg}=1\left(mol\right)\)
\(n_{O2}=0,25\left(mol\right)\)
\(PTHH:2Mg+O_2\rightarrow2MgO\)
_______0,1_____0,05____0,1 mol
a. Chất còn dư là O2
\(n_{O2\left(dư\right)}=n_{O2\left(bđ\right)}-n_{O2\left(pu\right)}=0,25-0,05=0,2\left(mol\right)\)
\(\Rightarrow m_{O2\left(Dư\right)}=6,4\left(g\right)\)
b. \(m_{MgO}=0,1.40=4\left(g\right)\)