Part A
How much potassium chlorate is needed to produce 22.0 mL of oxygen gas at 675. mmHg and 22. ∘C?
Express the mass to three significant digits.
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?g
Part B
If oxygen gas were collected over water at 22 ∘C and the total pressure of the wet gas were 710 mmHg , what would be the partial pressure of the oxygen?
Express the partial pressule in mm Hg to three significant digits
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?mmHg |
Part C
An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen. If 23.0 g of this gas were found to occupy 5.6 L at STP, what are the empirical and molecular formulas for this oxide of nitrogen?
Express the empirical and molecular formulas for this oxide of nitrogen, separated by a comma.
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Part A
Balanced equation:
2 KClO3(s) ====> 2 KCl(s) + 3
O2(g)
Reaction type: decomposition.
Let us calculate the moles of O2
PV = nRT
P = 665 mm Hg = 0.875 atm
V = 0.022 Liter
T = 273 + 22 = 295 K
R = 0.0821 L atm K-1 Mol-1
n = 0.875 x 0.022 / 0.0821 x 295 = 7.9481 x 10-4 Mole
Moles of KClO3 = 0.00052987 Moles
Mass of KClO3 = 0.00052987 x 122.54 = 0.0649 gm
Part B
the partial pressure of the oxygen = 675 mmHg
Paritial pressure of water = 710 -675 = 35 mm Hg
Part C
Let us calculate the moles of gas
PV = nRT
P = 1 atm V = 5.6 L
n = ? T = 273 K R = 0.0821 L atm K-1 Mol-1
n = 1 x 5.6 / 273 x 0.0821 = 0.2498 Moles
Molecular weight = 23 gm / 0.2498 = 92.05 g/mol
Empirical Formula of the compound = N2O4
Molar mass of N2O4 = 92.0110 g/mol
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