Question

What volume of oxygen gas, collected by displacement of water and measured at 749.0 tore and 30.0 c (vapor pressure of water 30.0 c =31.8 torr) will be formed by the decomposition of 11.0 g of potassium chlorate?

Answer 1

2KClO3$\to$ 2KCl+3O2

$m(O2)=\dfrac{11.0g*3*32g/mol}{122.55g/mol*2}=4.31g$

$V=m*R*T/M*P=\dfrac{4.31*62.364L*torr/К^{-1}*mol^{-1}*303.15К}{32g/mol*717.2torr}=3.55L$