Potassium Chlorate decomposes according to the reaction below. 2KClO3(s)  2KCl(s) + 3O2(g)

A 4.35 g sample of KClO3 is heated and the O2 gas produced by the reaction is collected in an evacuated flask. What is the volume of the O2 gas if the pressure of the flask is 0.75 atm and the gas temperature is 27oC? R=0.0821 (L*atm)/(mol*K)

1.75L

Explanation:

Reaction of decomposition is:

2KClO₃(s) →  2KCl(s) + 3O₂(g)

We determine moles of salt:

4.35 g . 1 mol /122.55 g = 0.0355 moles

Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen

Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.

We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:

We convert T° from °C to K → 27°C + 273 = 300K

P . V = n . R . T

0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K

V = (0.0532 mol  . 0.0821 L.atm/mol.K . 300K) / 0.75 atm

V  = 1.75 Liters

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol