Question

Answer 1

\(PTHH:2Mg+O_2->2MgO\)

                 0,2      0,1          0,2

\(n_{MgO}=\dfrac{m}{M}=\dfrac{8}{24+16}=0,2\left(mol\right)\)

\(=>n_{Mg}=0,2\cdot2:2=0,2\left(mol\right)\)

\(m_{Mg}=n\cdot M=0,2\cdot24=4,8\left(g\right)\)

\(n_{O_2}=0,2\cdot1:2=0,1\left(mol\right)\)

\(m_{O_2}=n\cdot M=0,1\cdot32=3,2\left(g\right)\)