Phản ứng xảy ra:
\(2Mg+O_2\rightarrow2MgO\)
Ta có:
\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(n_{O2}=\frac{5,6}{22,4}=0,25\left(mol\right)>\frac{1}{2}n_{Mg}\)
Nên O2 dư
\(\Rightarrow n_{O2\left(dư\right)}=0,25-\frac{1}{2}n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{O2}=0,2.32=6,4\left(g\right)\)
\(n_{MgO}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,1.40=4\left(g\right)\)