Question

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.

?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

?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.

Answer 1

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