Answer:
The right answer is "236.53 g".
Explanation:
The given values are:
P = 1.09 atm
V = 186 liters
The reaction will be:
⇒ [tex]KClO_3 (s)\rightarrow 2Kcl(s)+3O_2(g)[/tex]
The moles of O₂ will be:
= [tex]\frac{PV}{RT}[/tex]
On substituting the values, we get
= [tex]\frac{1.09\times 186}{0.0821\times 850}[/tex]
= [tex]\frac{202.47}{69.785}[/tex]
= [tex]2.90 \ moles[/tex]
Now,
1 mole O₂ is produced from
= [tex]\frac{2}{3} \ mol \ KClO_3[/tex]
then,
2.90 mole O₂ is produced from 2 mol KClO₃
= [tex]\frac{2}{3}\times 2.90[/tex]
= [tex]1.93 \ mol \ KClO_3[/tex]
hence,
The number of grans of solid in the engine will be:
= [tex]1.93\times 122.55[/tex]
= [tex]236.53 \ g[/tex]
