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Question 11.11: How many liters of oxygen gas form when 294 g of KClO3 compl......

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.

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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)

294  \cancel{g  KClO_3} \times \frac{1  \cancel{mol  KClO_3}}{122.5  \cancel{g  KClO_3}} \times \frac{3  mol  O_2}{2  \cancel{mol  KClO_3}}=3.60  mol  O_2

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

sol 11.11

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