Answer to Question #56153 in General Chemistry for sue
When heated, KClO3 solid forms solid KCl and O2 gas. A sample of KClO3 is heated and 365 mL of gas with a pressure of 790 mmHg is collected over water, at 26 ∘C. At 26 ∘C, the vapor pressure of water is 25 mmHg: 2(KClO3)(s)→2(KCl)(s)+3(O2)(g)
Question: How many grams of KClO3 were reacted?
Express your answer with the appropriate units.
m = n * M
PV = nRT,
so n = PV / RT
T(KClO3) = 26 ∘C + 273 = 299 K
P(O2) = P(total) – P(water) = 790 mm Hg – 25 mm Hg = 765 mm Hg
n (O2) = 765 mmHg * 365 mL / 63360 (mm Hg/ mol * K) / 299 K = 279225 / 1894464 = 0.15 mol
n (KClO3) = n (O2)/ 1.5 = 0.15 mol / 1.5 = 0.01 mol
m (KClO[sub]3[/sub) = 0.01 mol * 122.5 g/mol = 12 g
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