Đáp án:
$n_{O_2}=\dfrac{48}{32}=1,5(mol)$
a. $2KClO_3\to 2KCl + 3O_2$
b. $n_{KClO_3}=\dfrac{2}{3}\times n_{O_2}=\dfrac 23\times 1,5=1(mol)$
$\to m_{KClO_3}=1.122,5=122,5(g)$
c. $4P+5O_2\to 2P_2O_5$
$n_{P_2O_5}=\dfrac{2}{5}.n_{O_2}=\dfrac 25.1,5=0,6(mol)$
$\to m_{P_2O_5}=0,6.142=85,2(g)$