How many liters of oxygen gas form when 294 g of KClO_3 completely react in this reaction (which is used in the ignition of fireworks)?
2 KClO_3(s) → 2 KCl(s) + 3 O_2(g)Assume that the oxygen gas is collected at P = 755 mm Hg and T = 305 K.
GIVEN:
294 g KClO_3
P = 755 mm Hg (of oxygen gas)
T = 305 K
FIND: Volume of O_2 in liters
RELATIONSHIPS USED
1 mol KClO_3 = 122.6 g (molar mass of KClO_3)
2 mol KClO_3 : 3 mol O_2 (from balanced equation given in problem)
PV = nRT (ideal gas law, Section 11.8)
PV = nRT
V\,=\,\ \frac{n R T}{P} \\ P=755 \cancel{mm Hg} \times \frac{1 atm}{760 \cancel{mm Hg}}=0.99\underline{3}42 atm \\ V=\frac{3.60 \cancel{mol} \times 0.0821 \frac{L \cdot \cancel{atm}}{\cancel{mol} \cdot \cancel{K}} \times 305 \cancel{K}}{0.99\underline{3}42 \cancel{atm}}= 90.7 L