Answer :
We are given that heating 1 kg of pure KClO produces [tex]$k=399$[/tex] liters of oxygen at STPL. To find the total volume of oxygen produced when heating 245 kg of KClO, we use the relationship
[tex]$$
\text{Volume of oxygen} = (\text{mass of KClO}) \times (\text{conversion factor}).
$$[/tex]
Substituting the given values:
[tex]$$
\text{Volume of oxygen} = 245 \, \text{kg} \times 399 \, \frac{\text{liters}}{\text{kg}}.
$$[/tex]
Performing the multiplication:
[tex]$$
\text{Volume of oxygen} = 245 \times 399 = 97755 \, \text{liters}.
$$[/tex]
Thus, the volume of oxygen produced by heating 245 kg of pure KClO at STPL is
[tex]$$
\boxed{97755 \, \text{liters}}.
$$[/tex]
[tex]$$
\text{Volume of oxygen} = (\text{mass of KClO}) \times (\text{conversion factor}).
$$[/tex]
Substituting the given values:
[tex]$$
\text{Volume of oxygen} = 245 \, \text{kg} \times 399 \, \frac{\text{liters}}{\text{kg}}.
$$[/tex]
Performing the multiplication:
[tex]$$
\text{Volume of oxygen} = 245 \times 399 = 97755 \, \text{liters}.
$$[/tex]
Thus, the volume of oxygen produced by heating 245 kg of pure KClO at STPL is
[tex]$$
\boxed{97755 \, \text{liters}}.
$$[/tex]
