2Mg+O2\(\overset{t^0}{\rightarrow}2MgO\)
-Áp dụng định luật BTKl ta có:
\(m_{Mg}+m_{O_2}=m_{MgO}=8\)
\(\rightarrow\)\(1,5m_{O_2}+m_{O_2}=8\rightarrow2,5m_{O_2}=8\)
\(\rightarrow\)\(m_{O_2}=\dfrac{8}{2,5}=3,2gam\)
\(\rightarrow\)\(m_{Mg}=1,5m_{O_2}=1,5.3,2=4,8gam\)