PV = nRT
P = [(763-28.3)/760] atm.
V = 0.605 is correct.
Substitute R and T and solve for n = number of mols O2.
Convert mols O2 to mols KClO3 using the coefficients in the balanced equation. Then convert mols KClO3 to grams. g = mols x molar mass.
Then %KClO3 = (mass KClO3/2.677)*100 = ?
You are given 2.677 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl
and O2,
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),
and 605 mL of O2 is collected over water at 28 °C. The total pressure of the gases in the collection flask is
763 mm Hg. What is the weight percentage of KClO3 in the sample?
The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 28 °C is 28.3 mm Hg
(763-28.3)(0.605)/(0.0821)....help?
3 answers
n= PV/RT
[(763-28.3)/760](0.605)/(8.314)(301.15)
=0.000233592 x (2/3) = 0.000155728 x 122.5 g
=0.019076722/2.677 x 100 = 0.710 %
This result seems really small...what have i done incorrect?
[(763-28.3)/760](0.605)/(8.314)(301.15)
=0.000233592 x (2/3) = 0.000155728 x 122.5 g
=0.019076722/2.677 x 100 = 0.710 %
This result seems really small...what have i done incorrect?
Yes, and it IS too small.
R = 0.08206 and not 8.314 when P is in atmospheres. Correct that and see if the answer is about 72%
R = 0.08206 and not 8.314 when P is in atmospheres. Correct that and see if the answer is about 72%
2 answers