Em tham khảo nha :
\(\begin{array}{l}
d)\\
2C{H_3}OH + 3{O_2} \to 2C{O_2} + 4{H_2}O\\
19)\\
a)\\
10KI + 2KMn{O_4} + 8{H_2}S{O_4} \to 5{I_2} + 2MnS{O_4} + 6{K_2}S{O_4} + 8{H_2}O\\
{m_{KI}} = \dfrac{{20}}{{166}} = 0,12mol\\
{n_{{I_2}}} = \dfrac{{{n_{KI}}}}{2} = 0,06mol\\
{m_{{I_2}}} = 0,06 \times 254 = 15,24g\\
{n_{KMn{O_4}}} = \dfrac{{{n_{KI}}}}{5} = 0,024mol\\
{V_{KMn{O_4}}} = \dfrac{{0,024}}{2} = 0,012l\\
{n_{{K_2}S{O_4}}} = \dfrac{6}{{10}}{n_{KI}} = 0,072mol\\
{C_{{M_{{K_2}S{O_4}}}}} = \dfrac{{0,072}}{{0,012}} = 6M
\end{array}\)
