Đáp án:

\(V=1120\\ m=1,92\)

Giải thích các bước giải:

\(2KMnO_4\xrightarrow{t^{\circ}} K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{15,8}{158}=0,1\text{(mol)}\to n_{O_2}=\dfrac{0,1}2=0,05\text{(mol)}\to \begin{cases}80\%n_{O_2}=0,05\times 80\%=0,04\text{(mol)}\\V=0,05\times 22,4=1,12\text{(l)}=1120\text{(ml)}\end{cases}\\ Mg+\dfrac 12O_2\xrightarrow{t^{\circ}} MgO\\\to n_{Mg}=0,04\times 2=0,08\text{(mol)}\to m=24\times 0,08=1,92\text{(g)}\)

$n_{KMnO_4}=15,8/158=0,1mol$

$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

 Theo pt :

$n_{O_2}=1/2.n_{KMnO_4}=1/2.0,1=0,05mol$

$⇒V=0,05.22,4=1,12l=1120ml$

$80\%n_{O_2}=80\%.0,05=0,04mol$

$2Mg+O_2\overset{t^o}\to 2MgO$

Theo pt :

$n_{Mg}=2.n_{O_2}=2.0,04=0,08mol$

$⇒m_{Mg}=0,08.24=1,92g$