Đáp án:

 

Giải thích các bước giải:

$$n_{KMnO_4}=\dfrac{15,8}{158}=0,1(mol)$$

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

Theo PT: $n_{O_2}=0,5n_{KMnO_4}=0,05(mol)$$

$$\to n_{O_2(tt)}=0,05.75\%=0,0375(mol)$$

$$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$$

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

Đặt $n_{Mg}=x(mol)$

$$\to n_{Al}=2x(mol)$$

Theo PT: $\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}=n_{O_2}$

$$\to \dfrac{3}{4}.2x+\dfrac{1}{2}x=0,0375$$

$$\to x=0,01875(mol)$$

Theo PT: $n_{MgO}=x=0,01875(mol);n_{Al_2O_3}=x=0,01875(mol)$

$$\to m_{\rm oxit}=0,01875(40+102)=2,6625(g)$$

Mình gửi bạn.

#$HuyHoang0710$