\(n_{Mg}=\frac{6}{24}=0,25\left(mol\right)\)
\(n_{O2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
a.\(PTHH:2Mg+O_2\rightarrow2MgO\)
Trước ___0,25___0,1______
Phản ứng__0,2 ___0,1___________
Sau ____ 0,0___ 0 ______ 0,2
\(\rightarrow\) Mg dư, \(m_{Mg}=0,05.24=1,2\left(g\right)\)
b. \(n_{MgO}=0,2\left(mol\right)\)
\(\rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c. \(n_{O2}=2,24\left(l\right)\)