$\text{Đáp án+Giải thích các bước giải:}$

$\text{A.}$

$\text{2Zn + $O_{2}$$\xrightarrow{t^o}$2ZnO}$

$\text{B.}$

$\text{Ta có: $n_{Zn}$=$\dfrac{6,5}{65}$= 0,1 (mol)}$

$\text{Theo PT ta có: $n_{O_{2}}$=$\dfrac{1}{2}$$n_{Zn}$=$\dfrac{1}{2}$ × 0,1= 0,05 (mol)}$

$\text{⇒ $V_{O_{2}}$=0,05 × 22,4= 1,12 (lít)}$

$\text{C.}$

$\text{$2KMnO_{4}$$\xrightarrow{t^o}$$K_{2}MnO_{4}$ + $MnO_{2}$ + $O_{2}$}$

$\text{Ta có: $n_{O_{2}}$= 0,05 (mol)}$

$\text{Theo PT ta có: $n_{KMnO_{4}}$=2$n_{O_{2}}$=2 × 0,05= 0,1 (mol)}$

$\text{⇒ $m_{KMnO_{4}}$=0,1 × 158= 15,8 (g)}$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
b)\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
{n_{ZnO}} = {n_{Zn}} = 0,1\,mol\\
{m_{ZnO}} = 0,1 \times 81 = 8,1g\\
{n_{{O_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{O_2}}} = 0,05 \times 22,4 = 1,12l\\
c)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,1\,mol\\
{m_{KMn{O_4}}} = 0,1 \times 158 = 15,8g
\end{array}\)